vor|ed lxemp|esto lurocode ? Vo|ume ! A cement end concrete |ndustry pu|||cet|on |or tho dosign of in-situ concroto o|omonts in framod bui|dings to 8S £N 1992-1-1: 2004 and its Uk Nationa| Annox: 2005 CH Coodchi|d bSc Clng MClOb MlStructl ot a| This publication is available for purchase from www.concretecentre.com 2 |oroword The introduction of European standards to UK construction is a significant event as, for the first time, all design and construction codes within the EU will be harmonised. The ten design standards, known as the Eurocodes, will affect all design and construction activities as all current British Standards for structural design are due to be withdrawn in 2010. The cement and concrete industry recognised the need to enable UK design professionals to use Eurocode 2, Design of concrete structures, quickly, effectively, efficiently and with confidence. Supported by government, consultants and relevant industry bodies, the Concrete Industry Eurocode 2 Group (CIEG) was formed in 1999 and this Group has provided the guidance for a coordinated and collaborative approach to the introduction of Eurocode 2. As a result, a range of resources are being delivered by the concrete sector (see www.eurocode2.info). The aim of this publication, Worked Examples to Eurocode 2: Volume 1 is to distil from Eurocode 2, other Eurocodes and other sources the material that is commonly used in the design of concrete framed buildings. These worked examples are published in two parts. Volume 2 will include chapters on Foundations, Serviceability, Fire and Retaining walls. Acknow|odgomonts The original ideas for this publication emanates from the research project `Eurocode 2: Transition from UK to European concrete design standards’, which was led by the BCA and part funded by the DTI under their PII scheme and was overseen by a Steering Group and the CIEG. The work has been brought to fruition by The Concrete Centre from early initial drafts by various authors listed on the inside back cover. The concrete industry acknowledges and appreciates the support given by many individuals, companies and organisations in the preparation of this document. These are listed on the inside back cover. We gratefully acknowledge the authors of the initial drafts and the help and advice given by Robin Whittle in checking the text. Thanks are also due to Gillian Bond, Kevin Smith, Sally Huish and the design team at Michael Burbridge Ltd for their work on the production. The copyright of British Standards extracts reproduced in this document is held by the British Standards Institution (BSI). Permission to reproduce extracts from British Standards is granted by BSI under the terms of Licence No: 2009RM010. No other use of this material is permitted. This publication is not intended to be a replacement for the standard and may not reflect the most up-to-date status of the standard. British Standards can be obtained in PDF or hard copy formats from the BSI online shop: http://shop.bsigroup.com or by contacting BSI Customer Services for hard copies only: Tel:+44 (0)20 8996 9001, Email:
[email protected]. Þub|ishod by 1ho Concroto Contro, part of tho Minora| Þroducts Association k|vers|de louse, 4 Meedows bus|ness ler|, Stet|on Approech, b|ec|weter, Cem|er|ey, Surrey C0!/ 9Ab 1o|: ÷44 (0)!?/6 606S00 |ax: ÷44 (0)!?/6 606S0! www.concrotocontro.com ¹he Concrete Centre |s pert o¦ the M|nere| lroducts Assoc|et|on, the trede essoc|et|on ¦or the eggregetes, esphe|t, cement, concrete, ||me, morter end s|||ce send |ndustr|es www.minora|products.org Cement end Concrete lndustry lu|||cet|ons (CCll) ere produced through en |ndustry |n|t|et|ve to pu|||sh techn|ce| gu|dence |n support o¦ concrete des|gn end construct|on CCll pu|||cet|ons ere eve||e||e ¦rom the Concrete boo|shop et www.concrotobookshop.com 1o|: ÷44 (0)/004-60//// CCll-04! lu|||shed Lecem|er ?009 lSbN 9/S-!-9046S!S-S4-/ lr|ce Croup l © MlA - ¹he Concrete Centre All advice or information from MPA - The Concrete Centre is intended only for use in the UK by those who will evaluate the significance and limitations of its contents and take responsibility for its use and application. No liability (including that for negligence) for any loss resulting from such advice or information is accepted by MPA - The Concrete Centre or its subcontractors, suppliers or advisors. Readers should note that the publications from MPA - The Concrete Centre are subject to revision from time to time and should therefore ensure that they are in possession of the latest version. Printed by Michael Burbridge Ltd, Maidenhead, UK. This publication is available for purchase from www.concretecentre.com i Contents vor|ed lxemp|es to lurocode ? Vo|ume ! Symbo|s || 1 |ntroduction ! !! A|m ! !? lurocode bes|s o¦ structure| des|gn 3 !3 lurocode ! Act|ons on structures 4 !4 lurocode ? Les|gn o¦ concrete structures 4 !S Net|one| Annexes S !6 bes|s o¦ the wor|ed exemp|es |n th|s pu|||cet|on S !/ Assumpt|ons 6 !S Meter|e| propert|es 6 !9 lxecut|on 6 2 Ana|ysis, actions and |oad arrangomonts / ?! Methods o¦ ene|ys|s / ?? Act|ons / ?3 Cherecter|st|c ve|ues o¦ ect|ons / ?4 Ver|e||e ect|ons |mposed |oeds S ?S Ver|e||e ect|ons snow |oeds !? ?6 Ver|e||e ect|ons w|nd |oeds !3 ?/ Ver|e||e ect|ons others !/ ?S lermenent ect|ons !S ?9 Les|gn ve|ues o¦ ect|ons ?! ?!0 loed errengement o¦ ect|ons |ntroduct|on ?S ?!! loed errengements eccord|ng to the 0l Net|one| Annex to lurocode ?S ?!? lxemp|es o¦ |oed|ng ?/ 3 S|abs 3S 30 Cenere| 3S 3! S|mp|y supported one-wey s|e| 36 3? Cont|nuous one-wey so||d s|e| 40 33 Cont|nuous r|||ed s|e| S? 34 l|et s|e| /! 3S Ste|r ¦||ght 9S 4 8oams 9/ 40 Cenere| 9/ 4! Cont|nuous |eem on p|n supports 9S 4? leev||y |oeded l-|eem !04 43 Cont|nuous w|de ¹-|eem !!9 5 Co|umns !34 S0 Cenere| !34 S! ldge co|umn !3S S? ler|meter co|umn (|nterne| env|ronment) !39 S3 lnterne| co|umn !46 S4 Sme|| per|meter co|umn su||ect to two-hour ¦|re res|stence !S/ 6 wa||s !66 60 Cenere| !66 6! Sheer we|| !6/ 7 koforoncos and furthor roading !S3 Appondix A: Dorivod formu|ao !SS Appondix 8: Sorvicoabi|ity |imit stato !90 Appondix C: Dosign aids !94 Vo|umo 2 (pu|||shed seperete|y) loundet|ons Spec|e| dete||s strut 8 t|e Serv|cee||||ty l|re kete|n|ng we|| ke¦erences end ¦urther reed|ng Append|x A Ler|ved ¦ormu|ee Append|x b Les|gn e|ds This publication is available for purchase from www.concretecentre.com ii Symbols and abbreviations used in this publication Symbol Definition / Cross-sect|one| eree, Acc|dente| ect|on / S|te e|t|tude, m (snow) / A|t|tude o¦ the s|te |n metres e|ove meen see |eve| (w|nd) /, S, C Ver|e||es used |n the determ|net|on o¦ l ||m / c Cross-sect|one| eree o¦ concrete / d Les|gn ve|ue o¦ en ecc|dente| ect|on / ld Les|gn ve|ue o¦ e se|sm|c ect|on / re¦ ke¦erence eree o¦ the structure or structure| e|ement (w|nd) / s Cross-sect|one| eree o¦ re|n¦orcement / s,m|n M|n|mum cross-sect|one| eree o¦ re|n¦orcement / s,prov Aree o¦ stee| prov|ded / s,req Aree o¦ stee| requ|red / s! Aree o¦ re|n¦orc|ng stee| |n |eyer ! / s? Aree o¦ compress|on stee| (|n |eyer ?) / s| Aree o¦ the tens||e re|n¦orcement extend|ng et |eest / |d ÷ J |eyond the sect|on cons|dered / sM (/ sN ) ¹ote| eree o¦ re|n¦orcement requ|red |n symmetr|ce|, rectengu|er co|umns to res|st moment (ex|e| |oed) us|ng s|mp||¦|ed ce|cu|et|on method / sw Cross-sect|one| eree o¦ sheer re|n¦orcement, Aree o¦ punch|ng sheer re|n¦orcement |n one per|meter eround the co|umn / sw,m|n M|n|mum cross-sect|one| eree o¦ sheer re|n¦orcement, M|n|mum eree o¦ punch|ng sheer re|n¦orcement |n one per|meter eround the co|umn / t Aree o¦ tens||e re|n¦orcement |n ¦|et s|e| co|umn str|ps o L|stence, e||owence et supports o Ax|s d|stence ¦rom the concrete sur¦ece to the centre o¦ the |er (¦|re) o An exponent (|n cons|der|ng ||ex|e| |end|ng o¦ co|umns) o lro|ect|on o¦ the ¦oot|ng ¦rom the ¦ece o¦ the co|umn or we|| o | L|stence |y wh|ch the |ocet|on where e |er |s no |onger requ|red ¦or |end|ng moment |s d|sp|eced to e||ow ¦or the ¦orces ¦rom the truss mode| ¦or sheer ('Sh|¦t' d|stence ¦or curte||ment) o ! , o ? , L|stence ¦rom edge o¦ support to centre o¦ support o ! , / ! L|mens|ons o¦ the contro| per|meter eround en e|ongeted support (punch|ng sheer) ems| A|t|tude e|ove meen see |eve| / Overe|| w|dth o¦ e cross-sect|on, or ¦|enge w|dth |n e ¹- or l-|eem / breedth o¦ |u||d|ng (w|nd) / e l¦¦ect|ve w|dth o¦ e ¦|et s|e| (ed|ecent to per|meter co|umn) / e¦¦ l¦¦ect|ve w|dth o¦ e ¦|enge / eq (/ eq ) lqu|ve|ent w|dth (he|ght) o¦ co|umn /(/) ¦or rectengu|er sect|ons / m|n M|n|mum w|dth o¦ we| on ¹-, l- or l-|eems / t Meen w|dth o¦ the tens|on .one lor e ¹-|eem w|th the ¦|enge |n compress|on, on|y the w|dth o¦ the we| |s te|en |nto eccount / w v|dth o¦ the we| on ¹-, l- or l-|eems M|n|mum w|dth |etween tens|on end compress|on chords / ! le|¦ o¦ d|stence |etween ed|ecent we|s o¦ downstend |eems C e lxposure coe¦¦|c|ent (snow) C t ¹herme| coe¦¦|c|ent (snow) C w Sheer centre This publication is available for purchase from www.concretecentre.com Symbol Definition c ! , c ? L|mens|ons o¦ e rectengu|er co|umn lor edge co|umns, c ! |s meesured perpend|cu|er to the ¦ree edge (punch|ng sheer) c e|t A|t|tude ¦ector (w|nd) c d Lynem|c ¦ector (w|nd) c d|r L|rect|one| ¦ector (w|nd) c e,¦|et lxposure ¦ector (w|nd) c ¦ lorce coe¦¦|c|ent (w|nd) c m|n M|n|mum cover, (due to the requ|rements ¦or |ond, c m|n,| or dure||||ty c m|n,dur ) c nom Nom|ne| cover Nom|ne| cover shou|d set|s¦y the m|n|mum requ|rements o¦ |ond, dure||||ty end ¦|re c pe (lxterne|) pressure coe¦¦|c|ent (w|nd) c pe,!0 (lxterne|) pressure coe¦¦|c|ent ¦or erees > ! m ? (w|nd) c p| lnterne| pressure coe¦¦|c|ent (w|nd) c pro| lro|e||||ty ¦ector (w|nd) c seeson Seeson ¦ector (w|nd) c s S|.e ¦ector (w|nd) c y , c x Co|umn d|mens|ons |n p|en Dc dev A||owence mede |n des|gn ¦or dev|et|on / L|emeter o¦ e c|rcu|er co|umn, L|emeter J l¦¦ect|ve depth to tens|on stee| J ? l¦¦ect|ve depth to compress|on stee| J c l¦¦ect|ve depth o¦ concrete |n compress|on / l¦¦ect o¦ ect|on, lntegr|ty (|n ¦|re), l|est|c modu|us / cd Les|gn ve|ue o¦ modu|us o¦ e|est|c|ty o¦ concrete / cm Secent modu|us o¦ e|est|c|ty o¦ concrete /I bend|ng st|¦¦ness / s Les|gn ve|ue o¦ modu|us o¦ e|est|c|ty o¦ re|n¦orc|ng stee| lxp lxpress|on lQ0 Stet|c equ||||r|um e lccentr|c|ty e 0 M|n|mum eccentr|c|ty |n co|umns e ? Le¦|ect|on (used |n essess|ng / ? |n s|ender co|umns) e | lccentr|c|ty due to |mper¦ect|ons e y , e . lccentr|c|ty, / ld /\ ld e|ong , end · exes respect|ve|y (punch|ng sheer) l! lector to eccount ¦or ¦|enged sect|ons (de¦|ect|on) l? lector to eccount ¦or |r|tt|e pert|t|ons |n essoc|et|on w|th |ong spens (de¦|ect|on) l3 lector to eccount ¦or serv|ce stress |n tens||e re|n¦orcement (de¦|ect|on) / Act|on llM l|xed end moment / c (/ s ) lorce |n concrete (stee|) / d Les|gn ve|ue o¦ en ect|on / l ¹ens||e ¦orce |n re|n¦orcement to |e enchored / | Cherecter|st|c ve|ue o¦ en ect|on / rep kepresentet|ve ect|on ( y/ | where y ¦ector to convert cherecter|st|c to representet|ve ect|on) / s ¹ens||e ¦orce |n re|n¦orcement / td Les|gn ve|ue o¦ the tens||e ¦orce |n |ong|tud|ne| re|n¦orcement D/ td Add|t|one| tens||e ¦orce |n |ong|tud|ne| re|n¦orcement due to the truss sheer mode| iii This publication is available for purchase from www.concretecentre.com Symbol Definition / V,ld ¹ote| vert|ce| |oed (on |reced end |rec|ng mem|ers) / w kesu|tent cherecter|st|c ¦orce due to w|nd (See sect|on ?6) | |d 0|t|mete |ond stress | cd Les|gn ve|ue o¦ concrete compress|ve strength | c| Cherecter|st|c compress|ve cy||nder strength o¦ concrete et ?S deys | ct,d Les|gn tens||e strength o¦ concrete (a ct | ct,| /g C ) | ct,| Cherecter|st|c ex|e| tens||e strength o¦ concrete | ctm Meen ve|ue o¦ ex|e| tens||e strength o¦ concrete | sc Compress|ve stress |n compress|on re|n¦orcement et 0lS | yd Les|gn y|e|d strength o¦ |ong|tud|ne| re|n¦orcement, / s| | y| Cherecter|st|c y|e|d strength o¦ re|n¦orcement | ywd Les|gn y|e|d strength o¦ the sheer re|n¦orcement | ywd,e¦ l¦¦ect|ve des|gn strength o¦ punch|ng sheer re|n¦orcement | yw| Cherecter|st|c y|e|d strength o¦ sheer re|n¦orcement O | Cherecter|st|c ve|ue o¦ e permenent ect|on O |,sup 0pper cherecter|st|c ve|ue o¦ e permenent ect|on O |,|n¦ lower cherecter|st|c ve|ue o¦ e permenent ect|on q | Cherecter|st|c ve|ue o¦ e permenent ect|on per un|t |ength or eree / | lor|.onte| ect|on epp||ed et e |eve| / le|ght o¦ |u||d|ng (w|nd) / Overe|| depth o¦ e cross-sect|on, le|ght / eve O|struct|on he|ght (w|nd) / d|s L|sp|ecement he|ght (w|nd) / ¦ Lepth o¦ ¦oot|ng, ¹h|c|ness o¦ ¦|enge / s Lepth o¦ s|e| I Second moment o¦ eree o¦ concrete sect|on, lnert|e I lnsu|et|on (|n ¦|re) / ked|us o¦ gyret|on / / ld //J ? | c| A meesure o¦ the re|et|ve compress|ve stress |n e mem|er |n ¦|exure / lector to eccount ¦or structure| system (de¦|ect|on) / Ve|ue o¦ / e|ove wh|ch compress|on re|n¦orcement |s requ|red / v A correct|on ¦ector ¦or ex|e| |oed / f A correct|on ¦ector ¦or creep / Coe¦¦|c|ent or ¦ector / ke|et|ve ¦|ex|||||ty or re|et|ve st|¦¦ness / C|eer he|ght o¦ co|umn |etween end restre|nts / le|ght o¦ the structure |n metres / (or /) length, Spen / 0 l¦¦ect|ve |ength (o¦ co|umns) / 0 L|stence |etween po|nts o¦ .ero moment / 0 Les|gn |ep |ength / 0,¦| l¦¦ect|ve |ength under ¦|re cond|t|ons / | bes|c enchorege |ength / |d Les|gn enchorege |ength / |,eq lqu|ve|ent enchorege |ength / |,m|n M|n|mum enchorege |ength iv This publication is available for purchase from www.concretecentre.com Symbol Definition / |,rqd bes|c enchorege |ength / e¦¦ l¦¦ect|ve spen / n C|eer spen / y , / . Spens o¦ e two-wey s|e| |n the , end · d|rect|ons / bend|ng moment Moment ¦rom ¦|rst order ene|ys|s / Moment cepec|ty o¦ e s|ng|y re|n¦orced sect|on (e|ove wh|ch compress|on re|n¦orcement |s requ|red) / 0,lqp l|rst order |end|ng moment |n ques| permenent |oed com||net|on (SlS) / 0! , / 0? l|rst order end moments et 0lS /nc/aJ/nq e||owences ¦or |mper¦ect|ons / 0ld lqu|ve|ent ¦|rst order moment |nc|ud|ng the e¦¦ect o¦ |mper¦ect|ons (et e|out m|d he|ght) / 0ld,¦| l|rst order moment under ¦|re cond|t|ons / ? Nom|ne| second order moment |n s|ender co|umns / ld Les|gn ve|ue o¦ the epp||ed |nterne| |end|ng moment / ldy , / ld. Les|gn moment |n the respect|ve d|rect|on / kdy , / kd. Moment res|stence |n the respect|ve d|rect|on / t Les|gn trens¦er moment to co|umn ¦rom e ¦|et s|e| m Num|er o¦ vert|ce| mem|ers contr||ut|ng to en e¦¦ect m Mess / Ax|e| ¦orce / bes|c spen-to-e¦¦ect|ve-depth ret|o, //J, ¦or / !0 / 0ld,¦| Ax|e| |oed under ¦|re cond|t|ons NA Net|one| Annex / e , / | long|tud|ne| ¦orces contr||ut|ng to / | / ld Les|gn ve|ue o¦ the epp||ed ex|e| ¦orce (tens|on or compress|on) et 0lS NLl Net|one||y Leterm|ned leremeter(s) es pu|||shed |n e country's Net|one| Annex n loed |eve| et norme| temperetures Conservet|ve|y n 0/ (¦|re) n Ax|e| stress et 0lS n 0|t|mete ect|on (|oed) per un|t |ength (or eree) n ke|et|ve ex|e| ¦orce / ld /(/ c | cd ) n |e| ¹he ve|ue o¦ n et mex|mum moment res|stence n 0 , n s Num|er o¦ storeys Ç | Cherecter|st|c ve|ue o¦ e ver|e||e ect|on Ç |! (Ç || ) Cherecter|st|c ve|ue o¦ e |eed|ng ver|e||e ect|on (Cherecter|st|c ve|ue o¦ en eccompeny|ng ver|e||e ect|on) ¡ | Cherecter|st|c ve|ue o¦ e ver|e||e ect|on per un|t |ength or eree ¡ | bes|c w|nd pressure ¡ p lee| w|nd pressure ¡ p (· e ) lee| ve|oc|ty pressure et re¦erence he|ght · e , (w|nd) / kes|stence, Mechen|ce| res|stence (|n ¦|re) / A keect|on et support A / b keect|on et support b / d Les|gn ve|ue o¦ the res|stence to en ect|on · ked|us · m ket|o o¦ ¦|rst order end moments |n co|umns et 0lS SlS Serv|cee||||ty ||m|t stete(s) ÷ correspond|ng to cond|t|ons |eyond wh|ch spec|¦|ed serv|ce requ|rements ere no |onger met · Spec|ng · Snow |oed on e roo¦ v This publication is available for purchase from www.concretecentre.com Symbol Definition · | Cherecter|st|c ground snow |oed · r ked|e| spec|ng o¦ per|meters o¦ sheer re|n¦orcement · t ¹engent|e| spec|ng sheer re|n¦orcement e|ong per|meters o¦ sheer re|n¦orcement | ¹h|c|ness, ¹|me |e|ng cons|dered, breedth o¦ support | 0 ¹he ege o¦ concrete et the t|me o¦ |oed|ng 0lS 0|t|mete ||m|t stete(s) ÷ essoc|eted w|th co||epse or other ¦orms o¦ structure| ¦e||ure a ler|meter o¦ concrete cross-sect|on, hev|ng eree / c a ler|meter o¦ thet pert wh|ch |s exposed to dry|ng a C|rcum¦erence o¦ outer edge o¦ e¦¦ect|ve cross-sect|on (tors|on) a 0 ler|meter ed|ecent to co|umns (punch|ng sheer) a ! bes|c contro| per|meter (et ?J ¦rom ¦ece o¦ |oed) (punch|ng sheer) a ! ´ keduced contro| per|meter et per|meter co|umns (et ?J ¦rom ¦ece o¦ |oed) (punch|ng sheer) a | length o¦ the contro| per|meter under cons|deret|on (punch|ng sheer) a out ler|meter et wh|ch sheer re|n¦orcement |s no |onger requ|red \ Sheer ¦orce \ ld Les|gn ve|ue o¦ the epp||ed sheer ¦orce \ kd,c Sheer res|stence o¦ e mem|er w|thout sheer re|n¦orcement \ kd,mex Sheer res|stence o¦ e mem|er ||m|ted |y the crush|ng o¦ compress|on struts \ kd,cm|n M|n|mum sheer res|stence o¦ mem|er cons|der|ng concrete e|one \ kd,s Sheer res|stence o¦ e mem|er governed |y the y|e|d|ng o¦ sheer re|n¦orcement . | bes|c w|nd ve|oc|ty . |,0 ¹he ¦undemente| |es|c w|nd ve|oc|ty |e|ng the cherecter|st|c !0 m|nute w|nd ve|oc|ty et !0 m e|ove ground |eve| |n open country . |,mep lundemente| |es|c w|nd ve|oc|ty ¦rom l|gure NA! m/s . ld lunch|ng sheer stress . ld Sheer stress ¦or sect|ons u/|/oa| sheer re|n¦orcement ( \ ld // w J) . ld,. Sheer stress ¦or sect|ons w|th sheer re|n¦orcement ( \ ld // w · \ ld // w 09J) . kd,c Les|gn sheer res|stence o¦ concrete w|thout sheer re|n¦orcement expressed es e stress . kd,mex Cepec|ty o¦ concrete struts expressed es e stress w ! lector correspond|ng to e d|str||ut|on o¦ sheer (punch|ng sheer) w e lee| externe| w|nd |oed w | Cherecter|st|c ve|ue o¦ w|nd ect|on (Nb not |n the lurocodes end shou|d |e regerded es e ¦orm o¦ Ç | , cherecter|st|c ve|ue o¦ e ver|e||e ect|on) u | Cherecter|st|c un|t w|nd |oed u | Crec| w|dth u mex l|m|t|ng ce|cu|eted crec| w|dth `0, `A, `C Concrete exposure c|esses `L, `l, `S \ Neutre| ex|s depth \ L|stence |etween |u||d|ngs (w|nd) \ L|stence o¦ the sect|on |e|ng cons|dered ¦rom the centre ||ne o¦ the support \, ,, · Co-ord|netes, l|enes under cons|deret|on \ u Lepth o¦ the neutre| ex|s et the u|t|mete ||m|t stete e¦ter red|str||ut|on ´ ?one num|er o|te|ned ¦rom mep (snow) · lever erm o¦ |nterne| ¦orces vi This publication is available for purchase from www.concretecentre.com vii Symbol Definition · ke¦erence he|ght (w|nd) · e ke¦erence he|ght ¦or w|ndwerd we||s o¦ rectengu|er |u||d|ngs (w|nd) a Ang|e, Ang|e o¦ sheer ||n|s to the |ong|tud|ne| ex|s, ket|o a A A coe¦¦|c|ent ¦or use w|th e representet|ve ver|e||e ect|on te||ng |nto eccount eree supported a ! , a ? , a 3 lectors dee||ng w|th enchorege end |eps o¦ |ers a 4 , a S , a 6 a cc (a ct ) A coe¦¦|c|ent te||ng |nto eccount |ong term e¦¦ects o¦ compress|ve (tens||e) |oed end the wey |oed |s epp||ed a e Modu|er ret|o / s // cd a n A coe¦¦|c|ent ¦or use w|th e representet|ve ver|e||e ect|on te||ng |nto eccount num|er o¦ storeys supported b Ang|e, ket|o, Coe¦¦|c|ent b lector dee||ng w|th eccentr|c|ty (punch|ng sheer) g lert|e| ¦ector g C lert|e| ¦ector ¦or concrete g l lert|e| ¦ector ¦or ect|ons, / g C lert|e| ¦ector ¦or permenent ect|ons, O g C|,sup lert|e| ¦ector to |e epp||ed to O |,|n¦ g C|,|n¦ lert|e| ¦ector to |e epp||ed to O |,sup g Q lert|e| ¦ector ¦or ver|e||e ect|ons, Ç g M lert|e| ¦ector ¦or meter|e| (usue||y g C or g S ) g S lert|e| ¦ector ¦or re|n¦orc|ng stee| d ked|str||ut|on ret|o eque|s ret|o o¦ the red|str||uted moment to the e|est|c |end|ng moment (! ÷ % red|str||ut|on) e cu 0|t|mete compress|ve stre|n |n the concrete e cu? 0|t|mete compress|ve stre|n ||m|t |n concrete wh|ch |s not ¦u||y |n pure ex|e| compress|on essum|ng use o¦ the pere|o||c÷rectengu|er stress÷stre|n re|et|onsh|p (numer|ce||y e cu? e cu3 ) e cu3 0|t|mete compress|ve stre|n ||m|t |n concrete wh|ch |s not ¦u||y |n pure ex|e| compress|on essum|ng use o¦ the ||||neer stress÷stre|n re|et|onsh|p e sc Compress|ve stre|n |n re|n¦orcement e st ¹ens||e stre|n |n re|n¦orcement n lector de¦|n|ng e¦¦ect|ve strength ( ! ¦or < CS0/60) n ! Coe¦¦|c|ent ¦or |ond cond|t|ons n ? Coe¦¦|c|ent ¦or |er d|emeter y Ang|e, Ang|e o¦ compress|on struts (sheer) y | lnc||net|on used to represent |mper¦ect|ons l S|enderness ret|o l lector de¦|n|ng the he|ght o¦ the compress|on .one ( 0S ¦or < CS0/60) l ¦| S|enderness |n ¦|re l ||m l|m|t|ng s|enderness ret|o (o¦ co|umns) m | , m ! , m ? Snow |oed shepe ¦ectors m ¦| ket|o o¦ the des|gn ex|e| |oed under ¦|re cond|t|ons to the des|gn res|stence o¦ the co|umn et norme| tempereture |ut w|th en eccentr|c|ty epp||ce||e to ¦|re cond|t|ons v Strength reduct|on ¦ector ¦or concrete crec|ed |n sheer j keduct|on ¦ector/d|str||ut|on coe¦¦|c|ent lector epp||ed to O | |n bS lN !990 lxp (6!0|) r kequ|red tens|on re|n¦orcement ret|o r Lens|ty o¦ e|r (w|nd) r ke|n¦orcement ret|o ¦or requ|red compress|on re|n¦orcement, / s? //J This publication is available for purchase from www.concretecentre.com viii Symbol Definition r ! lercentege o¦ re|n¦orcement |epped w|th|n 06S/ 0 ¦rom the centre ||ne o¦ the |ep |e|ng cons|dered r | ke|n¦orcement ret|o ¦or |ong|tud|ne| re|n¦orcement r |y, r |. ke|n¦orcement ret|o o¦ |onded stee| |n the , end · d|rect|on r 0 ke¦erence re|n¦orcement ret|o | c| 0S !0 ÷3 s gd Les|gn ve|ue o¦ the ground pressure s s Stress |n re|n¦orcement et SlS s s A|so|ute ve|ue o¦ the mex|mum stress perm|tted |n the re|n¦orcement |mmed|ete|y e¦ter the ¦ormet|on o¦ the crec| s sc (s st ) Stress |n compress|on (end tens|on) re|n¦orcement s sd Les|gn stress |n the |er et the u|t|mete ||m|t stete s su 0nmod|¦|ed serv|ce stress |n re|n¦orcement determ|ned ¦rom 0lS |oeds (See l|gure C3) h(e,| 0 ) l|ne| ve|ue o¦ creep coe¦¦|c|ent h e¦ l¦¦ect|ve creep ¦ector f ber d|emeter c lectors de¦|n|ng representet|ve ve|ues o¦ ver|e||e ect|ons c 0 Com||net|on ve|ue o¦ e ver|e||e ect|on (eg used when cons|der|ng 0lS) c ! lrequent ve|ue o¦ e ver|e||e ect|on (eg used when cons|der|ng whether sect|on w||| heve crec|ed or not) c ? Ques|-permenent ve|ue o¦ e ver|e||e ect|on (eg used when cons|der|ng de¦ormet|on) w Mechen|ce| re|n¦orcement ret|o / s | yd // c | cd < ! This publication is available for purchase from www.concretecentre.com 1 Introduction Aim ¹he e|m o¦ th|s pu|||cet|on |s to |||ustrete through wor|ed exemp|es how bS lN !99?÷!÷! |!| (lurocode ?) mey |e used |n prect|ce to des|gn |n-s|tu concrete |u||d|ng structures lt |s |ntended thet these wor|ed exemp|es w||| exp|e|n how ce|cu|et|ons to lurocode ? mey |e per¦ormed lurocode ? str|ct|y cons|sts o¦ ¦our perts (lerts !÷!, !÷?, ? end 3) |!-4| |ut ¦or the purposes o¦ th|s pu|||cet|on, lurocode ? re¦ers to pert !÷! on|y, un|ess que||ued ¹he wor|ed exemp|es w||| |e cerr|ed out w|th|n the env|ronment o¦ other re|event pu|||cet|ons ||sted |e|ow, end |||ustreted |n l|gure !! ¹he other three perts o¦ lurocode ? N Other lurocodes N Meter|e| end execut|on stenderds N lu|||cet|ons |y the concrete |ndustry end others N ¹here ere, there¦ore, meny re¦erences to other documents end wh||e |t |s |ntended thet th|s pu|||cet|on, re¦erred to es wo·/eJ e\omo/e·, cen stend e|one, |t |s ent|c|peted thet users mey requ|re severe| o¦ the other re¦erences to hend, |n pert|cu|er, Conc/·e /a·ocoJe ? |S| , wh|ch summer|ses the ru|es end pr|nc|p|es thet w||| |e common|y used |n the des|gn o¦ re|n¦orced concrete ¦remed |u||d|ngs to lurocode ? BS EN 1991–1–1 BS EN 1992–1–1 DESIGN OF CONCRETE STRUCTURES BS EN 1991–2 ACTIONS NA NA NA NA NA NA NA NA NA NA NA –2 –3 –4 –6 PD6687 1–2 –3 –2 WORKED EXAMPLES TO EUROCODE 2 VOL 1 CONCRETE INDUSTRY PUBLICATIONS WORKED EXAMPLES PUBLICATIONS BY OTHERS Fire Bridges Liquid retaining Fire Snow Wind Execution Densities and imposed loads VOL 2 General STANDARDS PRECAST WORKED EXAMPLES BS EN 1990 BASIS OF DESIGN CONCISE EUROCODE 2 HOW TO DESIGN CONCRETE STRUCTURES www. Eurocode2 .info RC SPREAD SHEETS PRECAST DESIGN MANUAL MANUALS DETAILERS HANDBOOK DESIGN GUIDES BS EN 13670 EXECUTION OF CONCRETE STRUCTURES Noto ¹he term Stenderds encompesses br|t|sh Stenderds, lurocodes, Net|one| Annexes (NA) end lu|||shed Locuments |iguro 1.1 workod oxamp|os in contoxt ¹he des|gns ere |n eccordence w|th bS lN !99?÷!÷! |!| , es mod|ued |y the 0l Net|one| Annex |!e| end exp|e|ned |n lL 66S/ |6| lntroduct|on 1 1.1 This publication is available for purchase from www.concretecentre.com 2 Cenere||y, the ce|cu|et|ons ere cross-re¦erenced to the re|event c|euses |n e|| ¦our perts o¦ lurocode ? |!÷4| end, where eppropr|ete, to other documents See l|gure !? ¦or e gu|de to presentet|on ke¦erences to bS S!!0 |/| re¦er to lert ! un|ess otherw|se steted Cenere||y, the 's|mp|e' exemp|es depend on equet|ons end des|gn e|ds der|ved ¦rom lurocode ? ¹he der|ved equet|ons ere g|ven |n Append|x A end the des|gn e|ds ¦rom Sect|on !S o¦ Conc/·e /a·ocoJe ? |S| ere repeeted |n Append|x b ¹he exemp|es ere |ntended to |e eppropr|ete ¦or the|r purpose, wh|ch |s to |||ustrete the use o¦ lurocode ? ¦or |n-s|tu concrete |u||d|ng structures ¹here ere s|mp|e exemp|es to |||ustrete how typ|ce| hend ce|cu|et|ons m|ght |e done us|ng eve||e||e cherts end te||es der|ved ¦rom the Code ¹hese ere ¦o||owed |y more dete||ed exemp|es |||ustret|ng the dete||ed wor||ngs o¦ the Codes ln order to exp|e|n the use o¦ lurocode ?, severe| o¦ the ce|cu|et|ons ere presented |n dete|| ¦er |n excess o¦ thet necessery |n des|gn ce|cu|et|ons once users ere ¦em|||er w|th the Code ¹o en extent, the des|gns ere contr|ved to show ve||d methods o¦ des|gn|ng e|ements, to g|ve |ns|ght end to he|p |n ve||det|ng computer methods ¹hey ere not necesser||y the most eppropr|ete, the most econom|c or the on|y methods o¦ des|gn|ng the mem|ers |||ustreted Sections 1 and 2 Worked examples Cl. 6.4.4 ke|event c|euses or ¦|gure num|ers ¦rom bS lN !99?÷!÷! (|¦ the re¦erence |s to other perts, other lurocodes or other documents th|s w||| |e |nd|ceted) Cl. 6.4.4 NA lrom the re|event 0l Net|one| Annex (genere||y to bS lN !99?÷!÷!) NA Cl. 6.4.4 & NA lrom |oth bS lN !99?÷!÷! end 0l Net|one| Annex Cl. 6.4.4 & NA Fig. 2.1 Section 5.2 ke|event perts o¦ th|s pu|||cet|on Fig. 2.1 Section 5.2 EC1-1-1: 6.4.3 lrom other lurocodes bS lN !990, bS lN !99!, bS lN !99?÷!÷?, etc EC1-1-1: 6.4.3 PD 6687 [6] Soc/q·oanJ oooe· |o c/ /o|/ono/ /nne\e· S´ // '99?-' PD 6687 [6] Concise Conc/·e /a·ocoJe ? |S| Concise How to: Floors [8] /ou |o Je·/qn conc·e|e ·|·ac|a·e· a·/nq /a·ocoJe ? |S| //oo·· How to: Floors [8] Grey shaded tables ln Append|ces, der|ved content |n te||es not ¦rom lurocode ? |iguro 1.2 Cuido to prosontation As some o¦ the dete|||ng ru|es |n lurocode ? ere genere||y more |nvo|ved then those to bS S!!0, some o¦ the des|gns presented |n th|s pu|||cet|on heve |een extended |nto erees thet heve tred|t|one||y |een the respons|||||ty o¦ dete||ers ¹hese extended ce|cu|et|ons ere not necesser||y pert o¦ 'norme|' des|gn |ut ere |nc|uded et the end o¦ some ce|cu|et|ons lt |s essumed thet the des|gner w||| d|scuss end egree w|th the dete||er erees o¦ respons|||||ty end the degree o¦ ret|one||set|on, the extent o¦ des|gn|ng dete||s, essessment o¦ curte||ment end other espects thet the dete||er shou|d underte|e lt |s recogn|sed thet |n the vest me|or|ty o¦ ceses, the ru|es g|ven |n dete|||ng menue|s |S,9| w||| |e used lowever, the exemp|es ere |ntended to he|p when curte||ment, enchorege end |ep |engths need to |e determ|ned This publication is available for purchase from www.concretecentre.com 3 lntroduct|on Eurocode: Basis of structural design ln the lurocode system bS lN !990, lurocode So·/· o| ·|·ac|a·o/ Je·/qn |!0| overerches e|| the other lurocodes, bS lN !99! to bS lN !999 bS lN !990 deunes the e¦¦ects o¦ ect|ons, |nc|ud|ng geotechn|ce| end se|sm|c ect|ons, end epp||es to e|| structures |rrespect|ve o¦ the meter|e| o¦ construct|on ¹he meter|e| lurocodes deune how the e¦¦ects o¦ ect|ons ere res|sted |y g|v|ng ru|es ¦or des|gn end dete|||ng o¦ concrete, stee|, compos|te, t|m|er, mesonry end e|um|n|um (see l|gure !3) Structural safety, serviceability and durability Actions on structures Design and detailing Geotechnical and seismic design BS EN 1990, Eurocode: Basis of structural design BS EN 1991, Eurocode 1: Actions on structures BS EN 1992, Eurocode 2: Concrete BS EN 1993, Eurocode 3: Steel BS EN 1994, Eurocode 4: Composite BS EN 1995, Eurocode 5: Timber BS EN 1996, Eurocode 6: Masonry BS EN 1999, Eurocode 9: Aluminium BS EN 1997, Eurocode 7: Geotechnical design BS EN 1998, Eurocode 8: Seismic design |iguro 1.3 1ho £urocodo hiorarchy bS lN !990 prov|des the necessery |n¦ormet|on ¦or the ene|ys|s o¦ structures |nc|ud|ng pert|e| end other ¦ectors to |e epp||ed to the ect|ons ¦rom lurocode ! lt este|||shes the pr|nc|p|es end requ|rements ¦or the se¦ety, serv|cee||||ty end dure||||ty o¦ structures lt descr||es the |es|s ¦or des|gn es ¦o||ows A structure she|| |e des|gned end executed (constructed) |n such e wey thet |t w|||, dur|ng |ts |ntended ||¦e, w|th eppropr|ete degrees o¦ re||e||||ty end |n en econom|ce| wey Suste|n e|| ect|ons end |n¦|uences |||e|y to occur dur|ng execut|on end use N keme|n ¦|t ¦or the use ¦or wh|ch |t |s requ|red N ln other words, |t she|| |e des|gned us|ng ||m|t stetes pr|nc|p|es to heve edequete Ste||||ty N Structure| res|stence (|nc|ud|ng structure| res|stence |n ¦|re) N Serv|cee||||ty N Lure||||ty N lor |u||d|ng structures, e des|gn wor||ng ||¦e o¦ S0 yeers |s |mp||ed bS lN !990 stetes thet ||m|t stetes shou|d |e ver|ued |n e|| re|event des|gn s|tuet|ons pers|stent, trens|ent or ecc|dente| No re|event ||m|t stete she|| |e exceeded when des|gn ve|ues ¦or ect|ons end res|stences ere used |n des|gn ¹he ||m|t stetes ere 0|t|mete ||m|t stetes (0lS), wh|ch ere essoc|eted w|th co||epse or other ¦orms o¦ structure| N ¦e||ure Serv|cee||||ty ||m|t stetes (SlS), wh|ch correspond to cond|t|ons |eyond wh|ch spec|¦|ed N serv|ce requ|rements ere no |onger met A|| ect|ons ere essumed to very |n t|me end spece Stet|st|ce| pr|nc|p|es ere epp||ed to err|ve et the megn|tude o¦ the pert|e| |oed ¦ectors to |e used |n des|gn to ech|eve the requ|red re||e||||ty |ndex (|eve| o¦ se¦ety) ¹here |s en under|y|ng essumpt|on thet the ect|ons themse|ves ere descr||ed |n stet|st|ce| terms 1.2 EC0: 2.1 This publication is available for purchase from www.concretecentre.com 4 1.3 1.4 Eurocode 1: Actions on structures Act|ons ere deuned |n the !0 perts o¦ bS lN !99! lurocode ! /c|/on· on ·|·ac|a·e· |!!| bS lN !99!÷!÷! ?00? /en·/|/e· ·e/|-ue/q/| /moo·eJ /ooJ· |o· /a//J/nq· bS lN !99!÷!÷? ?00? /c|/on· on ·|·ac|a·e· e\oo·eJ |o |·e bS lN !99!÷!÷3 ?003 ´nou /ooJ· bS lN !99!÷!÷4 ?00S w/nJ oc|/on· bS lN !99!÷!÷S ?003 I/e·mo/ oc|/on· bS lN !99!÷!÷6 ?00S /c|/on· Ja·/nq e\eca|/on bS lN !99!÷!÷/ ?006 /cc/Jen|o/ oc|/on· bS lN !99!÷? ?003 /c|/on· on ·|·ac|a·e· I·o||c /ooJ· on /·/Jqe· bS lN !99!÷3 ?006 C·one· onJ moc//ne·, bS lN !99!÷4 ?006 ´//o· onJ |on/· ¹h|s pu|||cet|on |s me|n|y concerned w|th des|gn|ng ¦or the ect|ons deuned |y lert÷!÷! /en·/|/e· ·e/|-ue/q/| /moo·eJ /ooJ· |o· /a//J/nq· Les|gn ve|ues o¦ ect|ons end |oed errengements ere covered |n Sect|on ? Eurocode 2: Design of concrete structures lurocode ? /e·/qn o| conc·e|e ·|·ac|a·e· |!÷4| operetes w|th|n en env|ronment o¦ other luropeen end br|t|sh stenderds (see l|gure !3) lt |s governed |y bS lN !990 |!0| end su||ect to the ect|ons deuned |n lurocodes ! |!!| , / |!?| end S |!3| lt depends on ver|ous meter|e|s end execut|on stenderds end |s used es the |es|s o¦ other stenderds lert ?, S·/Jqe· |3| , end lert 3, //¡a/J ·e|o/n/nq onJ con|o/nmen| ·|·ac|a·e· |4| , wor| |y except|on to lert !÷! end !÷?, thet |s, c|euses |n lerts ? end 3 conurm, mod|¦y or rep|ece c|euses |n lert !÷! BS EN 1990 EUROCODE Basis of Structural Design BS EN 1991 EUROCODE 1 Basis of Structural Design BS EN 1992 EUROCODE 2 Design of concrete structures Part 1–1: General Rules for Structures Part 1–2: Structural Fire Design BS EN 1992 EUROCODE 2 Part 3: Liquid Retaining Structures BS EN 1995 EUROCODE 5 Design of Composite Structures BS EN 13670 Execution of Structures BS 8500 Specifying Concrete BS EN 206 Concrete BS EN 1992 EUROCODE 2 Part 2: Bridges BS EN 1997 EUROCODE 7 Geotechnical Design BS EN 1998 EUROCODE 8 Seismic Design BS EN 13369 Precast Concrete BS EN 10080 Reinforcing Steels BS 4449 Reinforcing Steels Noto lor c|er|ty Net|one| Annexes end exp|enetory documents (eg lL 66S/, end Non-Contred|ctory Comp|ementery ln¦ormet|on ÷ NCCl) ere not shown |iguro 1.4 £urocodo 2 in contoxt This publication is available for purchase from www.concretecentre.com 5 lntroduct|on 1.5 1.6 National Annexes lt |s the preroget|ve o¦ eech ClN Mem|er Stete to contro| |eve|s o¦ se¦ety |n thet country As e resu|t, some se¦ety ¦ectors end other peremeters |n the lurocodes, such es c||met|c cond|t|ons, dure||||ty c|esses end des|gn methods, ere su||ect to conurmet|on or se|ect|on et e net|one| |eve| ¹he dec|s|ons mede |y the net|one| |od|es |ecome Net|one||y Leterm|ned leremeters (NLls) wh|ch ere pu|||shed |n e Net|one| Annex (NA) ¦or eech pert o¦ eech lurocode ¹he Net|one| Annex mey e|so |nc|ude re¦erence to non-contred|ctory comp|ementery |n¦ormet|on (NCCl), such es net|one| stenderds or gu|dence documents ¹h|s pu|||cet|on |nc|udes re¦erences to the re|event Net|one| Annexes es eppropr|ete Basis of the worked examples in this publication ¹he des|gn ce|cu|et|ons |n th|s pu|||cet|on ere |n eccordence w|th bS lN !990, N /a·ocoJe So·/· o| ·|·ac|a·o/ Je·/qn |!0| end |ts 0l Net|one| Annex |!0e| bS lN !99!, lurocode ! N /c|/on· on ·|·ac|a·e· |n !0 perts |!!| end the|r 0l Net|one| Annexes |!!e| bS lN !99?÷!÷!, lurocode ? ÷ lert !÷! N /e·/qn o| conc·e|e ·|·ac|a·e· - Oene·o/ ·a/e· onJ ·a/e· |o· /a//J/nq· |!| end |ts 0l Net|one| Annex |!e| bS lN !99?÷!÷?, lurocode ? ÷ lert !÷? N /e·/qn o| conc·e|e ·|·ac|a·e· - Oene·o/ ·a/e· - ´|·ac|a·o/ |/·e Je·/qn |?| end |ts 0l Net|one| Annex |?e| lL 66S/, bec|ground peper to the N c/ /o|/ono/ /nne\e· |6| bS lN !99/, lurocode / N Oeo|ec/n/co/ Je·/qn - /o·| ' Oene·o/ ·a/e· |!?| end |ts 0l Net|one| Annex |!?e| ¹hey use meter|e|s con¦orm|ng to bS SS00÷! N Conc·e|e - Como/emen|o·, S·/|/·/ ´|onJo·J |o S´ // ?06-' /e|/oJ o| ·oec/|,/nq onJ qa/Jonce |o |/e ·oec/|/e· |!4| bS 4449 N ´|ee/ |o· |/e ·e/n|o·cemen| o| conc·e|e - we/Jo//e ·e/n|o·c/nq ·|ee/ - So· co// onJ Jeco//eJ o·oJac| - ´oec/|/co|/on |!S| ¹hey me|e re¦erence to severe| pu|||cet|ons, most note||y N Conc/·e /a·ocoJe ? |o· |/e Je·/qn o| /n-·/|a conc·e|e |·omeJ /a//J/nq· |o S´ // '99?-'-' ?004 onJ /|· c/ /o|/ono/ /nne\ ?00' |S| N /ou |o Je·/qn conc·e|e ·|·ac|a·e· a·/nq /a·ocoJe ? |S| ¹he execut|on o¦ the wor|s |s essumed to con¦orm to lL 66S/ N Soc/q·oanJ oooe· |o |/e c/ /o|/ono/ /nne\e· S´ // '99?-' |6| NSCS, N /o|/ono/ ·|·ac|a·o/ conc·e|e ·oec/|/co|/on |o· /a//J/nq con·|·ac|/on, 3rd ed|t|on |!6| Mey ?004 Or, when eve||e||e bS lN !36/0 N /\eca|/on o| conc·e|e ·|·ac|a·e· Lue ?0!0 |!/| As |mp|emented |y spec|¦|cet|ons such es NSCS, N /o|/ono/ ·|·ac|a·o/ conc·e|e ·oec/|/co|/on |o· /a//J/nq con·|·ac|/on, 4th ed|t|on |!S| CCll-0S0, due ?0!0 This publication is available for purchase from www.concretecentre.com 6 Assumptions Eurocode 2 lurocode ? essumes thet Les|gn end construct|on w||| |e underte|en |y eppropr|ete|y que||¦|ed end exper|enced personne| N Adequete superv|s|on end que||ty contro| w||| |e prov|ded N Meter|e|s end products w||| |e used es spec|¦|ed N ¹he structure w||| |e edequete|y me|nte|ned end w||| |e used |n eccordence w|th the des|gn |r|e¦ N ¹he requ|rements ¦or execut|on end wor|mensh|p g|ven |n lN !36/0 ere comp||ed w|th N The worked examples 0n|ess noted otherw|se, the ce|cu|et|ons |n th|s pu|||cet|on essume EC0: Table 2.1 A des|gn ||¦e o¦ S0 yeers N Table 3.1 ¹he use o¦ C30/3/ concrete N BS 4449 ¹he use o¦ Crede A, b or C re|n¦orcement, des|gneted 'l' |n eccordence w|th bS S666 N |!9| Table 4.1, BS 8500: Table A.1 lxposure c|ess `C! N Building Regs [20,21] ! hour ¦|re res|stence N Cenere||y eech ce|cu|et|on |s rounded end |t |s the rounded ve|ue thet |s used |n eny ¦urther ce|cu|et|on Material properties Meter|e| propert|es ere spec|ued |n terms o¦ the|r cherecter|st|c ve|ues ¹h|s usue||y corresponds to the |ower S% ¦rect||e o¦ en essumed stet|st|ce| d|str||ut|on o¦ the property cons|dered ¹he ve|ues o¦ g C end g S , pert|e| ¦ectors ¦or meter|e|s, ere |nd|ceted |n ¹e||e !! 1ab|o 1.1 Þartia| factors for matoria|s Dosign situation g C - concroto g S - roinforcing stoo| ULS – persistent and transient !S0 !!S Accidental – non-fire !?0 !00 Accidental – fire !00 !00 SLS !00 !00 Execution ln the 0l, LL lNV !36/0 |??| |s current|y eve||e||e |ut w|thout |ts Net|one| App||cet|on Locument lor |u||d|ng structures |n the 0l, the |ec|ground document lL 66S/ |6| cons|ders the prov|s|ons o¦ the Net|one| Structure| Concrete Spec|ucet|on (NSCS) |!6| to |e equ|ve|ent to those |n lN !36/0 ¦or to|erence c|ess ! vhen pu|||shed, bS lN !36/0 |!/| end, |¦ eppropr|ete, the correspond|ng Net|one| App||cet|on Locument w||| te|e precedence 1.7 1.7.1 1.7.2 1.8 1.9 Cl. 1.3 PD 6687 [6] Table 2.1 & NA This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 7 2 2.1 2.1.1 2.1.2 2.2 2.3 Analysis, actions and load arrangements Methods of analysis ULS At the u|t|mete ||m|t stete (0lS) the type o¦ ene|ys|s shou|d |e eppropr|ete to the pro||em |e|ng cons|dered ¹he ¦o||ow|ng ere common|y used l|neer e|est|c ene|ys|s N l|neer e|est|c ene|ys|s w|th ||m|ted red|str||ut|on N l|est|c ene|ys|s N Cl. 5.1.1(7) lor 0lS, the moments der|ved ¦rom e|est|c ene|ys|s mey |e red|str||uted prov|ded thet the resu|t|ng d|str||ut|on o¦ moments reme|ns |n equ||||r|um w|th the epp||ed ect|ons ln cont|nuous |eems or s|e|s w|th | c| < S0 Mle the m|n|mum e||owe||e ret|o o¦ the red|str||uted moment to the moment |n the ||neer ene|ys|s, d, |s 0/0 where C|ess b or C|ess C re|n¦orcement |s used or 0S0 where C|ess A re|n¦orcement |s used Cl. 5.5.4 & NA v|th|n the ||m|ts set, coe¦uc|ents ¦or moment end sheer der|ved ¦rom e|est|c ene|ys|s mey |e used to determ|ne ¦orces |n regu|er structures (see Append|x b) ¹he des|gn o¦ co|umns shou|d |e |esed on e|est|c moments w|thout red|str||ut|on Cl. 5.1.1 l|est|c ene|ys|s mey |e used ¦or des|gn et 0lS prov|ded thet the requ|red duct|||ty cen |e essured, ¦or exemp|e |y ||m|t|ng \ u /J (to < 0?S ¦or concrete strength c|esses < CS0/60), us|ng C|ess b or C re|n¦orcement, or ensur|ng the ret|o o¦ moments et |ntermed|ete supports to moments |n spens |s |etween 0S end ?0 Cl. 5.6.2 SLS At the serv|cee||||ty ||m|t stete (SlS) ||neer e|est|c ene|ys|s mey |e used l|neer e|est|c ene|ys|s mey |e cerr|ed out essum|ng Cross-sect|ons ere uncrec|ed end reme|n p|ene (|e ene|ys|s mey |e |esed on concrete N gross sect|ons) l|neer stress÷stre|n re|et|onsh|ps N ¹he use o¦ meen ve|ues o¦ e|est|c modu|us N Cl. 5.4(1) Actions Act|ons re¦er to |oeds epp||ed to the structure es deuned |e|ow lermenent ect|ons ere ect|ons ¦or wh|ch the ver|et|on |n megn|tude w|th t|me |s neg||g|||e N EC1-1-1: 2.1 Ver|e||e ect|ons ere ect|ons ¦or wh|ch the ver|et|on |n megn|tude w|th t|me |s not neg||g|||e N EC1-1-1: 2.2, 3.3.1(2) Acc|dente| ect|ons ere ect|ons o¦ short duret|on |ut o¦ s|gn|¦|cent megn|tude thet ere N un|||e|y to occur on e g|ven structure dur|ng the des|gn wor||ng ||¦e EC1-1-7 lmposed de¦ormet|ons ere not cons|dered |n th|s pu|||cet|on Characteristic values of actions ¹he ve|ues o¦ ect|ons g|ven |n the ver|ous perts o¦ lurocode ! /c|/on· on ·|·ac|a·e· |!!| ere te|en es cherecter|st|c ve|ues ¹he cherecter|st|c ve|ue o¦ en ect|on |s deuned |y one o¦ the ¦o||ow|ng three e|ternet|ves EC0: 4.1.2 lts meen ve|ue ÷ genere||y used ¦or permenent ect|ons N This publication is available for purchase from www.concretecentre.com 8 An upper ve|ue w|th en |ntended pro|e||||ty o¦ not |e|ng exceeded or |ower ve|ue w|th en N |ntended pro|e||||ty o¦ |e|ng ech|eved ÷ norme||y used ¦or ver|e||e ect|ons w|th |nown stet|st|ce| d|str||ut|ons, such es w|nd or snow A nom|ne| ve|ue ÷ used ¦or some ver|e||e end ecc|dente| ect|ons N Variable actions: imposed loads General lmposed |oeds on |u||d|ngs ere d|v|ded |nto cetegor|es ¹hose most ¦requent|y used |n concrete des|gn ere shown |n ¹e||e ?! EC1-1-1: Tables 6.1, 6.7, 6.9 & NA 1ab|o 2.1 Catogorios of imposod |oads Catogory Doscription A Arees ¦or domest|c end res|dent|e| ect|v|t|es B O¦¦|ce erees C Arees o¦ congreget|on D Shopp|ng erees E Storege erees end |ndustr|e| use (|nc|ud|ng eccess erees) F ¹re¦¦|c end per||ng erees (veh|c|es < 30 |N) G ¹re¦¦|c end per||ng erees (veh|c|es > 30 |N) H koo¦s (|neccess|||e except ¦or me|ntenence end repe|r) I koo¦s (eccess|||e w|th occupency cetegor|es A ÷ L) K koo¦s (eccess|||e ¦or spec|e| serv|ces, eg ¦or he||copter |end|ng erees) Notos 1 Cetegory ¦ |s not used 2 lor ¦or|||¦t |oed|ng re¦er to bS lN !99!÷!÷! C| 6?3 Characteristic values of imposed loads Cherecter|st|c ve|ues ¦or common|y used |mposed |oeds ere g|ven |n ¹e||es ?? to ?S EC1-1-1: Tables 6.1, 6.2 & NA.3 1ab|o 2.2 A: domostic and rosidontia| Sub- catogory £xamp|o |mposod |oads j k {kNIm 2 } J k {kN} A1 A|| useges w|th|n se|¦-conte|ned dwe|||ng un|ts Commune| erees (|nc|ud|ng ||tchens) |n sme|| a ||oc|s o¦ ¦|ets !S ?0 A2 bedrooms end dorm|tor|es, except those |n se|¦-conte|ned s|ng|e ¦em||y dwe|||ng un|ts end |n hote|s end mote|s !S ?0 A3 bedrooms |n hote|s end mote|s, hosp|te| werds, to||et erees ?0 ?0 A4 b||||erd/snoo|er rooms ?0 ?/ A5 be|con|es |n s|ng|e-¦em||y dwe|||ng un|ts end commune| erees |n sme|| a ||oc|s o¦ ¦|ets ?S ?0 A6 be|con|es |n hoste|s, guest houses, res|dent|e| c|u|s Commune| erees |n |erger a ||oc|s o¦ ¦|ets M|n 30 b M|n ?0 c A7 be|con|es |n hote|s end mote|s M|n 40 b M|n ?0 c Notos a Sme|| ||oc|s o¦ ¦|ets ere those w|th < 3 storeys end < 4 ¦|ets per ¦|oor/ste|rcese Otherw|se they ere cons|dered to |e |erger ||oc|s o¦ ¦|ets b Seme es the rooms to wh|ch they g|ve eccess, |ut w|th e m|n|mum o¦ 30 |N/m ? or 40 |N/m ? c Concentreted et the outer edge EC1-1-1: Tables 6.1, 6.2 & NA.3 1ab|o 2.3 8: officos Sub- catogory £xamp|o |mposod |oads j k {kNIm 2 } J k {kN} B1 Cenere| use other then |n b? ?S ?/ B2 At or |e|ow ground ¦|oor |eve| 30 ?/ 2.4 2.4.1 2.4.2 This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 9 1ab|o 2.4 C: aroas of congrogation Sub- catogory £xamp|o |mposod |oads j k J k C1 Aroas with tab|os C11 lu|||c, |nst|tut|one| end commune| d|n|ng rooms end |ounges, ce¦es end resteurents (Note use C4 or CS |¦ eppropr|ete) ?0 30 C12 keed|ng rooms w|th no |oo| storege ?S 40 C13 C|essrooms 30 30 C2 Aroas with fixod soats C21 Assem||y erees w|th ¦|xed seet|ng a 40 36 C22 l|eces o¦ worsh|p 30 ?/ C3 Aroas without obstac|os for moving poop|o C31 Corr|dors, he||weys, e|s|es |n |nst|tut|one| type |u||d|ngs, hoste|s, guest houses, res|dent|e| c|u|s end commune| erees |n |erger b ||oc|s o¦ ¦|ets 30 4S C32 Ste|rs, |end|ngs |n |nst|tut|one| type |u||d|ngs, hoste|s, guest houses, res|dent|e| c|u|s end commune| erees |n |erger b ||oc|s o¦ ¦|ets 30 40 C33 Corr|dors, he||weys, e|s|es |n other c |u||d|ngs 40 4S C34 Corr|dors, he||weys, e|s|es |n other c |u||d|ngs su||ected to whee|ed veh|c|es, |nc|ud|ng tro||eys S0 4S C35 Ste|rs, |end|ngs |n other c |u||d|ngs su||ected to crowds 40 40 C36 ve||weys ÷ l|ght duty (eccess su|te||e ¦or one person, we||wey w|dth epprox 600 mm) 30 ?0 C37 ve||weys ÷ Cenere| duty (regu|er two-wey pedestr|en tre¦¦|c) S0 36 C38 ve||weys ÷ leevy duty (h|gh-dens|ty pedestr|en tre¦¦|c |nc|ud|ng escepe routes) /S 4S C39 Museum ¦|oors end ert ge||er|es ¦or exh|||t|on purposes 40 4S C4 Aroas with possib|o physica| activitios C41 Lence he||s end stud|os, gymnes|e, steges d S0 36 C42 Lr||| he||s end dr||| rooms d S0 /0 C5 Aroas subjoctod to |argo crowds C51 Assem||y erees w|thout ¦|xed seet|ng, concert he||s, |ers end p|eces o¦ worsh|p d,e S0 36 C52 Steges |n pu|||c essem||y erees d /S 4S koy a l|xed seet|ng |s seet|ng where |ts remove| end the use o¦ the spece ¦or other purposes |s |mpro|e||e b Sme|| ||oc|s o¦ ¦|ets ere those w|th < 3 storeys end < 4 ¦|ets per ¦|oor/ste|rcese Otherw|se they ere cons|dered to |e '|erger' ||oc|s o¦ ¦|ets c Other |u||d|ngs |nc|ude those not covered |y C3! end C3?, end |nc|ude hote|s end mote|s end |nst|tut|one| |u||d|ngs su||ected to crowds d lor structures thet m|ght |e suscept|||e to resonence e¦¦ects, re¦erence shou|d |e mede to NA?! e lor grendstends end sted|e, re¦erence shou|d |e mede to the requ|rements o¦ the eppropr|ete cert|¦y|ng euthor|ty EC1-1-1: Tables 6.1, 6.2 & NA.3 This publication is available for purchase from www.concretecentre.com 10 EC1-1-1: Tables 6.1, 6.2 & NA.3 1ab|o 2.5 D: shopping aroas Sub- catogory £xamp|o |mposod |oads j k {kNIm 2 } J k {kN} D Shopping aroas D1 Arees |n genere| rete|| shops 40 36 D2 Arees |n depertment stores 40 36 EC1-1-1: Tables 6.3, 6.4 & NA.4, NA.5 1ab|o 2.6 £: storago aroas and industria| uso {inc|uding accoss aroas} Sub- catogory £xamp|o |mposod |oads j k {kNIm 2 } J k {kN} £1 Aroas suscoptib|o to accumu|ation of goods inc|uding accoss aroas E11 Cenere| erees ¦or stet|c equ|pment not spec|¦|ed e|sewhere (|nst|tut|one| end pu|||c |u||d|ngs) ?0 !S E12 keed|ng rooms w|th |oo| storege, eg |||rer|es 40 4S E13 Cenere| storege other then those spec|¦|ede ?4/m /0 E14 l||e rooms, ¦|||ng end storege spece (o¦¦|ces) S0 4S E15 Stec| rooms (|oo|s) ?4/m he|ght (m|n 6S) /0 E16 leper storege end stet|onery stores 40/m he|ght 90 E17 Lense mo|||e stec||ng (|oo|s) on mo|||e tro||eys |n pu|||c end |nst|tut|one| |u||d|ngs 4S/m he|ght /0 E18 Lense mo|||e stec||ng (|oo|s) on mo|||e truc|s |n werehouses 4S/m he|ght (m|n !S0) /0 E19 Co|d storege S0/m he|ght (m|n !S0) 90 £2 |ndustria| uso See bS lN !99!÷!÷! ¹e||es 6S 8 66 lor|||¦ts C|esses ll! to ll6 koy a lower |ound ve|ue g|ven More spec|¦|c |oed ve|ues shou|d |e egreed w|th c||ent 1ab|o 2.7 | and C: traffic and parking aroas Sub- catogory £xamp|o |mposod |oads j k {kNIm 2 } J k {kN} | 1raffic and parking aroas {vohic|os < 30 kN} ¹re¦¦|c end per||ng erees (veh|c|es < 30 |N) ?S S0 C 1raffic and parking aroas {vohic|os > 30 kN} ¹re¦¦|c end per||ng erees (veh|c|es > 30 |N) S0 ¹o |e determ|ned ¦or spec|¦|c use This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 11 2.4.3 1ab|o 2.8 H, | and k: roofs Sub- catogory £xamp|o |mposod |oads j k {kNIm 2 } J k {kN} H koofs {inaccossib|o oxcopt for maintonanco and ropair} koo¦ s|ope, a° < 30° 06 09 30° < a < 60° 06(60 ÷ a)/30 < 60° 0 | koofs {accossib|o with occupancy catogorios A - D} Cetegor|es A ÷ L As ¹e||es ?? to ?S eccord|ng to spec|¦|c use k koofs {accossib|o for spocia| sorvicos, o.g. for ho|icoptor |anding aroas} le||copter c|ess lC! (< ?0 |N) (su||ect to dynem|c ¦ector f !4) ÷÷ ?0 le||copter c|ess lC? (< 60 |N) ÷÷ 60 Notos 1 koo¦s ere cetegor|.ed eccord|ng to the|r eccess|||||ty lmposed |oeds ¦or roo¦s thet ere norme||y eccess|||e ere genere||y the seme es ¦or the spec|¦|c use end cetegory o¦ the ed|ecent eree lmposed |oeds ¦or roo¦s w|thout eccess ere g|ven e|ove 2 ¹here |s no cetegory ¦ EC1-1-1: 6.3.4.1(2), Tables 6.9, 6.10, 6.11 & NA.7 EC1-1-1: 6.3.4 & NA Movable partitions ¹he se|¦-we|ght o¦ move||e pert|t|ons mey |e te|en |nto eccount |y e un|¦orm|y d|str||uted |oed, ¡ | , wh|ch shou|d |e edded to the |mposed |oeds o¦ ¦oors es ¦o||ows EC1-1-1: 6.3.1.2 (8) & NA lor move||e pert|t|ons w|th e se|¦-we|ght o¦ !0 |N/m we|| |ength N ¡ | 0S |N/m ? lor move||e pert|t|ons w|th e se|¦-we|ght o¦ ?0 |N/m we|| |ength N ¡ | 0S |N/m ? lor move||e pert|t|ons w|th e se|¦-we|ght o¦ 30 |N/m we|| |ength N ¡ | !? |N/m ? leev|er pert|t|ons shou|d |e cons|dered seperete|y Reduction factors General koo¦s do not que||¦y ¦or |oed reduct|ons ¹he method g|ven |e|ow comp||es w|th the 0l Net|one| Annex |ut d|¦¦ers ¦rom thet g|ven |n the lurocode EC1-1-1: 6.3.1.2 (10) 6.3.1.2(11) & NA Area A reduct|on ¦ector ¦or |mposed |oeds ¦or eree, a A , mey |e used end shou|d |e determ|ned us|ng a A !0 ÷ A/!000 ~ 0/S where / |s the eree (m ? ) supported w|th |oeds que||¦y|ng ¦or reduct|on (|e cetegor|es A to l es ||sted |n ¹e||e ?!) EC1-1-1: 6.3.1.2 (10) & NA Exp. (NA.1) This publication is available for purchase from www.concretecentre.com 12 Number of storeys A reduct|on ¦ector ¦or num|er o¦ storeys, a n , mey |e used end shou|d |e determ|ned us|ng a n !! ÷ n/!0 a n 06 a n 0S ¦or ! < n < S ¦or S < n < !0 ¦or n > !0 where n num|er o¦ storeys w|th |oeds que||¦y|ng ¦or reduct|on (|e cetegor|es A to L es ||sted |n ¹e||e ?!) EC1-1-1: 6.3.1.2 (11) & NA Use Accord|ng to the 0l NA, a A end a n mey not |e used together Variable actions: snow loads EC1-1-3: 5.2(3) ln pers|stent or trens|ent s|tuet|ons, snow |oed on e roo¦, ·, |s deuned es |e|ng · m | C e C t · | where EC1-1-3: 5.3.1, 5.3.2 & NA m | snow |oed shepe ¦ector, , e|ther m ! or m ? m ! undr|¦ted snow shepe ¦ector m ? dr|¦ted snow shepe ¦ector lor ¦et roo¦s, 0° a (w|th no h|gher structures c|ose or e|utt|ng), m ! m ? 0S lor she||ow monop|tch roo¦s, 0°< a < 30° (w|th no h|gher structures c|ose or e|utt|ng), m ! 0S, m ? 0S (! ÷ a/30) lor other ¦orms o¦ roo¦ end |oce| e¦¦ects re¦er to bS lN !99!÷!÷3 Sect|ons S3 end 6 EC1-1-3: 5.2(7) &Table 5.1 C e exposure coe¦uc|ent lor w|ndswept topogrephy C e 0S lor norme| topogrephy C e !0 lor she|tered topogrephy C e !? EC1-1-3: 5.2(8) C | therme| coe¦uc|ent, C | !0 other then ¦or some g|ess-covered roo¦s, or s|m||er EC1-1-3: & NA 2.8 · / cherecter|st|c ground snow |oed |N/m ? 0!S(0!´ ÷ 00S) ÷ (/ ÷ !00)/S?S where ´ .one num|er o|te|ned ¦rom the mep |n bS lN !99!÷!÷3 NA l|gure NA! / s|te e|t|tude, m l|gure NA! o¦ the NA to bS lN !99!÷!÷3 e|so g|ves ugures ¦or · | et !00 m ems| essoc|eted w|th the .ones lor the me|or|ty o¦ the South lest, the M|d|ends, Northern lre|end end the north o¦ lng|end epert ¦rom h|gh ground, · | 0S0 |N/m ? lor the vest Country, vest ve|es end lre|end the ugure |s |ess lor most o¦ Scot|end end perts o¦ the eest coest o¦ lng|end, the ugure |s more See l|gure ?! Snow |oed |s c|ess|ued es e ver|e||e uxed ect|on lxcept|one| c|rcumstences mey |e treeted es ecc|dente| ect|ons |n wh|ch cese re¦erence shou|d |e mede to bS lN !99!÷!÷3 2.5 EC1-1-1: 6.3.1.2 (11) & NA Exp. ( NA.2) This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 13 1 1 1 Zone 1 = 0.25 kN/m 2 at 100 m a.m.s.l. Zone 1 = 0.40 kN/m 2 at 100 m a.m.s.l. Zone 1 = 0.50 kN/m 2 at 100 m a.m.s.l. Zone 1 = 0.60 kN/m 2 at 100 m a.m.s.l. Zone 1 = 0.70 kN/m 2 at 100 m a.m.s.l. 2 3 4 5 1 1 2 2 2 4 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 1 |iguro 2.1 Charactoristic ground snow |oad map {ground snow |oad at 100 m a.m.s.|. {kNIm 2 } EC1-1-3: NA Fig. NA.1 Variable actions: wind loads ¹h|s Sect|on presents e very s|mp|e |nterpretet|on o¦ lurocode ! |!!, !!e| end |s |ntended to prov|de e |es|c understend|ng w|th respect to rectengu|er-p|en |u||d|ngs w|th ¦et roo¦s ln genere|, mex|mum ve|ues ere g|ven w|th more |n¦ormet|on e |ower ve|ue m|ght |e used ¹he user shou|d |e cere¦u| to ensure thet eny |n¦ormet|on used |s w|th|n the scope o¦ the epp||cet|on env|seged ¹he user |s re¦erred to more spec|e||st gu|dence |?3, ?4| or bS lN !99!÷!÷4 |?S| end |ts 0l Net|one| Annex |?Se| ¹he Net|one| Annex |nc|udes c|eer end conc|se ¦ow cherts ¦or the determ|net|on o¦ pee| ve|oc|ty pressure, ¡ p ln essence cherecter|st|c w|nd |oed cen |e expressed es u | c ¦ ¡ p(.) where c ¦ ¦orce coe¦uc|ent, wh|ch ver|es, |ut |s e mex o¦ !3 ¦or overe|| |oed ¡ p(.) c e(.) c e¹ ¡ | where c e(.) exposure ¦ector ¦rom l|gure ?3 c e¹ town terre|n ¦ector ¦rom l|gure ?4 ¡ | 0006. | ? |N/m ? where . | . |,mep c e|t EC1-1-4: Figs NA.7, NA.8 where . |,mep ¦undemente| |es|c w|nd ve|oc|ty ¦rom l|gure ?? c e|t e|t|tude ¦ector, conservet|ve|y, c e|t ! ÷ 000!/ where / e|t|tude ems| EC1-1-4: Fig. NA.1 Sym|o|s e||rev|et|ons end some o¦ the ceveets ere exp|e|ned |n the sect|ons |e|ow, wh|ch together prov|de e procedure ¦or determ|n|ng w|nd |oed to bS lN !99!÷!÷4 2.6 This publication is available for purchase from www.concretecentre.com 14 Determine basic wind velocity, v b . | c d|r c seeson c pro| . |,0 EC1-1-4: 4.2(1) Note 2 & NA 2.4, 2.5 where c d|r d|rect|one| ¦ector Conservet|ve|y, c d|r !0 (c d|r |s e m|n|mum o¦ 0/3 or 0/4 ¦or w|nd |n en eester|y d|rect|on, 30° to !?0°) EC1-1-4: 4.2(2) Note 3 & NA 2.7: Fig. NA.2 c seeson seeson ¦ector lor e 6 month return per|od, |nc|ud|ng w|nter, or greeter, c seeson !00 EC1-1-4: 4.2(1) Notes 4 & 5 & NA 2.8 c pro| pro|e||||ty ¦ector !00 ¦or return per|od o¦ S0 yeers . |,0 . |,mep c e|t EC1-1-4: 4.2(1) Note 2 & NA 2.4: Fig. NA.1 where . |,mep ¦undemente| |es|c w|nd ve|oc|ty ¦rom l|gure ?? EC1-1-4: 4.2(2) Note 1 & NA 2.5 c e|t e|t|tude ¦ector Conservet|ve|y, c e|t ! ÷ 000!/ where / e|t|tude o¦ the s|te |n metres ems| vhere orogrephy |s s|gn|ucent (|e the s|te |s c|ose to e s|ope steeper then 00S), re¦er to NA ?S Calculate basic wind pressure, q b ¡ | 0Sr. | ? EC1-1-4: 4.5(1) Note 2 & NA 2.18 where . | es e|ove r dens|ty o¦ e|r !??6 |g/m 3 ( !?0 N/m 3 ) ¦or 0l Calculate peak wind pressure, q p(z) EC1-1-4: 4.5(1) Note 1 & NA 2.17 ¡ p(.) c e(.) ¡ | ¦or country |ocet|ons c e(.) c e¹ ¡ | ¦or town |ocet|ons where EC1-1-4: 4.5(1) Note 1, NA 2.17 & Fig. A.NA.1 ¡ | es e|ove c e(.) exposure ¦ector der|ved ¦rom l|gure ?3 et he|ght · (see |e|ow) c e,¹ exposure correct|on ¦ector ¦or town terre|n der|ved ¦rom l|gure ?4 · the he|ght et wh|ch ¡ p |s sought lor e w|ndwerd we|| end when / < /, ¡ p |s ce|cu|eted et the re¦erence he|ght · e / lor other espect ret|os // o¦ the w|ndwerd we||, ¡ p ,|s ce|cu|eted et d|¦¦erent re¦erence he|ghts ¦or eech pert (see bS lN !99!÷!÷4) where / he|ght o¦ |u||d|ng / |reedth o¦ |u||d|ng lor |eewerd end s|de we||s, · he|ght o¦ |u||d|ng 2.6.1 2.6.2 2.6.3 EC1-1-4: 4.5(1) Note 1 & NA 2.17: Fig. NA.8 EC1-1-4: 7.2.2(1), Note & NA 2.26 EC1-1-4: 4.5(1) Note 1 & NA 2.17: Fig. NA.7 This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 15 30 31 29 28 27 26 25 24 23 22 200 100 90 80 70 60 50 40 30 z - h d i s ( m ) 20 10 9 8 7 6 5 4 3 2 ≤0.1 1 10 Distance upwind to shoreline (km) ≥100 4.0 3.5 3.0 2.5 1.5 2.0 10 Use 1.0 in this area 0.9 0.8 0.7 200 100 90 80 70 60 50 40 30 z - h d i s ( m ) 20 10 9 8 7 6 5 4 3 2 ≤0.1 1 10 Distance inside town terrain (km) ≥20 Noto Subjoct to a|titudo corroction. Noto Conora||y a dis = 0. |or torrain catogory |V {towns otc.} soo 8S £N 1991-1-4: A.5. |iguro 2.2 Map of fundamonta| basic wind vo|ocity, o b,map , {mIs} |iguro 2.3 £xposuro factor \ o{z} for sitos in country or town torrain |iguro 2.4 Mu|tip|ior for oxposuro corroction for sitos in town torrain EC1-1-4: 4.2(1) Note 2 & NA 2.4: Fig. NA.1 EC1-1-4: 4.5(1) Note 1 & NA 2.17: Fig. NA.7 EC1-1-4: 4.5(1) Note 1 & NA 2.17: Fig. NA.8 Calculate characteristic wind load, w k u | ¡ p(.) c ¦ where ¡ p(.) es e|ove c ¦ ¦orce coe¦uc|ent ¦or the structure or structure| e|ement Cenere||y c pe ÷ c p| where EC1-1-4: 7, 8 & NA c pe (externe|) pressure coe¦uc|ent dependent on s|.e o¦ eree cons|dered end .one lor erees e|ove ! m ? , c pe,!0 shou|d |e used EC1-1-4: 7.2.1(1) Note 2 & NA. 2.25 Overall loads lor the we||s o¦ rectengu|er-p|en |u||d|ngs, c pe,!0 mey |e determ|ned ¦rom ¹e||e /! o¦ bS lN !99!÷!÷4 EC1-1-4: 7.2.2(2) Note 1 & NA.2.27 lowever, ¦or the determ|net|on o¦ overe|| |oeds on |u||d|ngs, the net pressure coe¦uc|ents g|ven |n ¹e||e ?9 mey |e used ln th|s cese |t w||| |e unnecessery to determ|ne |nterne| w|nd pressure coe¦uc|ents EC1-1-4: 7.2.2(2) Note 1 & NA.2.27, Table NA.4 Cladding loads lor erees e|ove ! m ? , c pe,!0 shou|d |e used c pe,!0 mey |e determ|ned ¦rom ¹e||e /! o¦ bS lN !99!÷!÷4 See ¹e||e ?!0 EC1-1-4: 7.2.2(2) Note 1 & NA.2.27 2.6.4 This publication is available for purchase from www.concretecentre.com 16 EC1-1-4: 7.2.3, NA.2.28 & NA advisory note Flat roofs lor ¦et roo¦s, eccord|ng to the Adv|sory Note |n the NA some o¦ the ve|ues o¦ c pe,!0 |n ¹e||e /? o¦ bS lN !99!÷!÷4 (see ¹e||e ?!!) ere s|gn|ucent|y d|¦¦erent ¦rom current prect|ce |n the 0l lt recommends thet des|gners shou|d cons|der us|ng the ve|ues |n bS 6399? to me|nte|n the current |eve|s o¦ se¦ety end economy See ¹e||e ?!? BS 6399: Table 8 & Fig. 18 EC1-1-4: NA.2.28 & NA advisory note lor other ¦orms o¦ roo¦ re¦er to bS lN !99!÷!÷4 end the 0l NA lt w||| e|so |e necessery to determ|ne |nterne| w|nd pressure coe¦uc|ents ¦or the des|gn o¦ c|edd|ng EC1-1-4: 7.2.9(6) Note 2 c p| |nterne| pressure coe¦uc|ent lor no dom|nent open|ngs c p| mey |e te|en es the more onerous o¦ ÷0? end ÷03 EC1-1-4: NA 2.27, Table NA.4 1ab|o 2.9 Not prossuro coofficiont, \ po,10 , for wa||s of roctangu|ar p|an bui|dings* a(] Not prossuro coofficiont, \ po,10 5 !3 1 !! ~ 0.25 0S Notos 1 ´ |n e¦¦ect these ve|ues ere ¦orce coe¦¦|c|ents ¦or determ|n|ng overe|| |oeds on |u||d|ngs 2 / he|ght o¦ |u||d|ng 3 / |reedth o¦ |u||d|ng (perpend|cu|er to w|nd) 4 J depth o¦ |u||d|ng (pere||e| to w|nd) 5 Ve|ues mey |e |nterpo|eted 6 lxc|udes ¦unne|||ng EC1-1-4: 7.2.2(2) Table 7.1, Note 1 & NA 2.27: Tables NA.4a , NA.4b 1ab|o 2.10 £xtorna| prossuro coofficiont, \ po,10 , for wa||s of roctangu|ar-p|an bui|dings Zono Doscription \ po,10 Max. Min. Zone A lor we||s pere||e| to the w|nd d|rect|on, erees w|th|n 0?m|n|/, ?/| o¦ w|ndwerd edge ÷!? Zone B lor we||s pere||e| to the w|nd d|rect|on, erees w|th|n 0?m|n|/, ?/| o¦ w|ndwerd edge ÷0S Zone C lor we||s pere||e| to the w|nd d|rect|on, erees ¦rom 0?m|n|/, ?/| to m|n|/, ?/| o¦ w|ndwerd edge ÷0S Zone D v|ndwerd we|| ÷0S Zone E leewerd we|| ÷0/ Zones D and E Net ÷!3 Notos 1 / he|ght o¦ |u||d|ng 2 / |reedth o¦ |u||d|ng (perpend|cu|er to w|nd) EC1-1-4: 7.2, Table 7.2 & NA 1ab|o 2.11 £xtorna| prossuro coofficiont, \ po,10 for f|at roofs* Zono Doscription \ po,10 Sharp odgo at oavos with parapot Zone F v|th|n 0!m|n||, ?h| o¦ w|ndwerd edge end w|th|n 0?m|n||, ?h| o¦ return edge (pere||e| to w|nd d|rect|on) ÷!S ÷!6 Zone G v|th|n 0!m|n||, ?h| o¦ w|ndwerd edge end outw|th 0?m|n||, ?h| o¦ return edge (pere||e| to w|nd d|rect|on) ÷!? ÷!! Zone H koo¦ |etween 0!m|n||, ?h| end 0Sm|n||, ?h| ¦rom w|ndwerd edge ÷0/ ÷0/ Zone I keme|nder |etween 0Sm|n||, ?h| end |eewerd edge ±0? ±0? Notos 1 ´ Accord|ng to NA to bS lN !99!-!-4, th|s te||e |s not recommended ¦or use |n the 0l 2 / he|ght o¦ |u||d|ng 3 / |reedth o¦ |u||d|ng (perpend|cu|er to w|nd) This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 17 1ab|o 2.12 £xtorna| prossuro coofficiont, \ po , for f|at roofs Zono Doscription \ po Sharp odgo at oavos with parapot Zone A v|th|n 0!m|n||, ?h| o¦ w|ndwerd edge end w|th|n 0?Sm|n||, ?h| o¦ return edge (pere||e| to w|nd d|rect|on) ÷?0 ÷!9 Zone B v|th|n 0!m|n||, ?h| o¦ w|ndwerd edge end outw|th 0?Sm|n||, ?h| o¦ return edge (pere||e| to w|nd d|rect|on) ÷!4 ÷!3 Zone C koo¦ |etween 0!m|n||, ?h| end 0Sm|n||, ?h| ¦rom w|ndwerd edge ÷0/ ÷0/ Zone D keme|nder |etween 0Sm|n||, ?h| end |eewerd edge ±0? ±0? Notos 1 / he|ght o¦ |u||d|ng 2 / |reedth o¦ |u||d|ng (perpend|cu|er to w|nd) EC1-1-4: 7.2.3, NA.2.28 & NA advisory note. BS 6399: Table 8 & Fig. 18 Calculate the overall wind force, F w / w c s c d Su | / re¦ where EC1-1-4: 5.3.2, Exp. (5.4) & NA u | es e|ove EC1-1-4: 6.2(1) a), 6.2(1) c) c s c d structure| ¦ector, conservet|ve|y !0 or mey |e der|ved EC1-1-4: 6.2(1) e) & NA.2.20 where c s s|.e ¦ector c s mey |e der|ved ¦rom lxp (6?) or te||e NA3 Lepend|ng on ve|ues o¦ (/ ÷ /) end (· ÷ / d|s ) end d|v|d|ng |nto ?one A, b or C, e ve|ue o¦ c s (e ¦ector < !00) mey |e ¦ound EC1-1-4: 6.3(1), Exp. (6.2) & NA.2.20, Table NA3 c d dynem|c ¦ector c d mey |e der|ved ¦rom lxp (63) or ugure NA9 Lepend|ng on ve|ues o¦ d s (|oger|thm|c decrement o¦ structure| demp|ng) end ///, e ve|ue o¦ c d (e ¦ector > !00) mey |e ¦ound EC1-1-4: 6.3(1), Exp. (6.3) & NA.2.20: Fig. NA9 c d mey |e te|en es !0 ¦or ¦remed |u||d|ngs w|th structure| we||s end mesonry |nterne| we||s, end ¦or c|edd|ng pene|s end e|ements / re¦ re¦erence eree o¦ the structure or structure| e|ement EC1-1-4: 5.3.2, Exp. (5.4) & NA Variable actions: others Act|ons due to construct|on, tre¦uc, ure, therme| ect|ons, use es s||os or ¦rom crenes ere outs|de the scope o¦ th|s pu|||cet|on end re¦erence shou|d |e mede to spec|e||st ||tereture EC1-1-6, EC1-2, EC1-1-2, EC1-1-5, EC1-3 & EC1-4 2.6.5 2.7 This publication is available for purchase from www.concretecentre.com 18 Permanent actions ¹he dens|t|es end eree |oeds o¦ common|y used meter|e|s, sheet meter|e|s end ¦orms o¦ construct|on ere g|ven |n ¹e||es ?!3 to ?!S Act|ons er|s|ng ¦rom sett|ement, de¦ormet|on end creep ere outs|de the scope o¦ th|s document |ut genere||y ere to |e cons|dered es permenent ect|ons vhere cr|t|ce|, re¦er to spec|e||st ||tereture 1ab|o 2.13 8u|k donsitios for soi|s and matoria|s [11, 26] 8u|k donsitios kNIm 3 8u|k donsitios kNIm 3 Soi|s Matoria|s C|ey ÷ st|¦¦ !9÷?? Concrete ÷ re|n¦orced ?S0 C|ey ÷ so¦t !6÷!9 Concrete ÷ wet re|n¦orced ?60 Crenu|er ÷ |oose !6÷!S C|ess ?S6 Crenu|er ÷ dense !9÷?! Cren|te ?/3 S||ty c|ey, sendy c|ey !6÷?0 lerdcore !90 Matoria|s l|mestone (lort|end stone ÷ med we|ght) ??0 Asphe|t ??S l|mestone (mer||e ÷ heevywe|ght) ?6/ b|oc|s ÷ eereted concrete (m|n) S0 Mecedem pev|ng ?!0 b|oc|s ÷ eereted concrete (mex) 90 MLl S0 b|oc|s ÷ dense eggregete ?00 l|ester !4! b|oc|s ÷ ||ghtwe|ght !40 l|ywood 63 boo|s ÷ |u|| storege S÷!! Sendstone ?3S br|c|wor| ÷ ||ue ?40 Screed ÷ send/cement ??0 br|c|wor| ÷ eng|neer|ng ??0 Stee|/|ron //0 br|c|wor| ÷ ¦|etton !S0 ¹errecotte ?0/ br|c|wor| ÷ london stoc| !90 ¹|m|er ÷ Loug|es ¦|r S? br|c|wor| ÷ send ||me ?!0 ¹|m|er ÷ luropeen |eech/oe| /! Ch|p|oerd 69 ¹|m|er ÷ Crede C!6 36 Concrete ÷ eereted !00 ¹|m|er ÷ Crede C?4 4! Concrete ÷ ||ghtwe|ght !S0 ¹|m|er ÷ lro|o/tee| 64 Concrete ÷ p|e|n ?40 2.8 This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 19 1ab|o 2.14 1ypica| aroa |oads for concroto s|abs and shoot matoria|s [11, 26] 1ypica| aroa |oads kNIm 2 1ypica| aroa |oads kNIm 2 Concroto s|abs Shoot matoria|s lrecest concrete so||d un|ts (!00 mm) ?S0 l|ester s||m coet 00S lrecest concrete ho||owcore un|ts a (!S0 mm) ?40 l|ester|oerd (!?S mm) 009 lrecest concrete ho||owcore un|ts a (?00 mm) ?S/ l|ester|oerd (!9 mm) 0!S lrecest concreteho||owcore un|ts a (300 mm) 40/ l|ywood (!?S mm) 00S lrecest concrete ho||owcore un|ts a (400 mm) 4S4 l|ywood (!9 mm) 0!? k|||ed s|e| b (?S0 mm) 400 Querry t||es |nc|ud|ng morter |edd|ng 03? k|||ed s|e| b (300 mm) 430 ke|sed ¦|oor ÷ heevy duty 0S0 k|||ed s|e| b (3S0 mm) 4/0 ke|sed ¦|oor ÷ med|um we|ght 040 ve¦¦|e s|e| c ÷ stenderd mou|ds (3?S mm) 600 ke|sed ¦|oor ÷ ||ghtwe|ght 030 ve¦¦|e s|e| c ÷ stenderd mou|ds (4?S mm) /30 kender (!3 mm) 030 ve¦¦|e s|e| c ÷ stenderd mou|ds (S?S mm) S60 Screed ÷ S0 mm !!S Shoot matoria|s Screed ÷ ||ghtwe|ght (?S mm) 04S Asphe|t (?0 mm) 046 Ste|n|ess stee| roo¦|ng (04 mm) 00S Cerpet end under|ey 00S Suspended ce|||ng ÷ stee| 0!0 Ch|p|oerd (!S mm) 0!? Suspended ¦||re|oerd t||es 00S Lry ||n|ng on stud (?0 mm) 0!S ¹8C |oerds (!SS mm) 009 le|se ce|||ng ÷ stee| ¦rem|ng 0!0 ¹8C |oerds (?? mm) 0!? le|t (3 |eyer) end ch|pp|ngs 03S ¹||es ÷ cerem|c ¦|oor on |edd|ng !00 C|ess ÷ dou||e g|e.|ng 0S? bettens ¦or s|et|ng end t|||ng 003 C|ess ÷ s|ng|e g|e.|ng 030 ¹||es ÷ c|ey roo¦ (mex) 06/ lnsu|et|on ÷ g|ess ¦||re (!S0 mm) 003 ¹||es ÷ neture| s|ete (th|c|) 06S lev|ng stones (S0 mm) !?0 ¹||es ÷ |nter|oc||ng concrete 0SS l|ester ÷ two coet gypsum (!? mm) 0?! ¹||es ÷ p|e|n concrete 0/S koy a lo||owcore ¦|gures essume no topp|ng (S0 mm structure| topp|ng : !?S |N/m ? ) b k|||ed s|e|s !S0 we| © /S0 centres w|th !00 mm th|c| ¦|enge/s|e| ve| s|ope !!0 c ve¦¦|e s|e|s !S0 r||s © 900 centres w|th !00 mm th|c| ¦|enge/s|e| ve| s|ope !!0 This publication is available for purchase from www.concretecentre.com 20 1ab|o 2.15 Loads for typica| forms of construction [26] Cavity wa|| {kNIm 2 } br|c|wor| !0?S mm ?40 lnsu|et|on S0 mm 00? b|oc|wor| !00 mm !40 l|ester 0?! 1ota| 4.0 Lightwoight c|adding {kNIm 2 } lnsu|eted pene| 0?0 lur||ns 00S Lry ||n|ng on stud 0!S 1ota| 0.40 Curtain wa||ing {kNIm 2 } A||ow !00 Þrocast concroto c|adding {kNIm 2 } lec|ng !00 lrecest pene| (!00 mm) ?40 lnsu|et|on 00S Lry ||n|ng on stud 0!S 1ota| 3.60 Dry |ining {kNIm 2 } Mete| studs 00S l|ester|oerd end s||m × ? 040 1ota| 0.45 1imbor stud wa|| {kNIm 2 } ¹|m|er studs 0!0 l|ester|oerd end s||m × ? 040 1ota| 0.50 Offico f|oor {kNIm 2 } Cerpet 003 ke|sed ¦|oor 030 Se|¦-we|ght o¦ ?S0 mm so||d s|e| 6?S Suspended ce|||ng 0!S Serv|ces 030 1ota| 7.03 Offico coro aroa {kNIm 2 } ¹||es end |edd|ng, e||ow !00 Screed ??0 Se|¦-we|ght o¦ ?S0 mm so||d s|e| 6?S Suspended ce|||ng 0!S Serv|ces 030 1ota| 9.90 Stairs {kNIm 2 } !S0 mm we|st ( z!/S © ?S |N/m 3 ) 440 ¹reeds 0!S × 0?S × 4/? © ?S |N/m 3 !SS Screed 00S © ?? |N/m 3 !!0 l|ester 0?! l|n|sh t||es 8 |edd|ng !00 1ota| 8.60 2.5c Loads for typica| forms of construction 1ab|o 2.14 2.5c Loads for typica| forms of construction kosidontia| f|oor {kNIm 2 } Cerpet 00S l|oet|ng ¦|oor 0!S Se|¦-we|ght o¦ ?S0 mm so||d s|e| 6?S Suspended ce|||ng 0?0 Serv|ces 0!0 1ota| 6.75 Schoo| f|oor {kNIm 2 } Cerpet/¦|oor|ng 00S Se|¦-we|ght o¦ ?S0 mm so||d s|e| 6?S Suspended ce|||ng 0!S Serv|ces 0?0 1ota| 6.60 Hospita| f|oor {kNIm 2 } l|oor|ng 00S Se|¦-we|ght o¦ ?S0 mm so||d s|e| 6?S Screed ??0 Suspended ce|||ng 0!S Serv|ces (|ut cen |e greeter) 00S 1ota| 8.70 ||at roofIoxtorna| torraco {kNIm 2 } lev|ng or greve|, e||ow ??0 veterproo¦|ng 0S0 lnsu|et|on 0!0 Se|¦-we|ght o¦ ?S0 mm so||d s|e| ce|||ng 6?S Suspended ce|||ng 0!S Serv|ces 030 1ota| 9.50 1imbor pitchod roof {kNIm 2 } ¹||es (renge 0S0÷0/S) 0/S bettens 00S le|t 00S ke¦ters 0!S lnsu|et|on 00S l|ester|oerd 8 s||m 0!S Serv|ces 0!0 Ce|||ng |o|sts 0!S 1ota| porpondicu|ar to roof 1.45 1ota| on p|an assuming 30° pitch 1.60 Mota| docking roof {kNIm 2 } lnsu|eted pene| 0?0 lur||ns 0!0 Stee|wor| 030 Serv|ces 0!0 1ota| 0.70 This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 21 Design values of actions General case ¹he des|gn ve|ue o¦ en ect|on, / d , thet occurs |n e |oed cese |s / d g l c / | where g l pert|e| ¦ector ¦or the ect|on eccord|ng to the ||m|t stete under cons|deret|on ¹e||e ?!6 |nd|cetes the pert|e| ¦ectors to |e used |n the 0l ¦or the com||net|ons o¦ representet|ve ect|ons |n |u||d|ng structures c / | mey |e cons|dered es the representet|ve ect|on, / rep , eppropr|ete to the ||m|t stete |e|ng cons|dered where c e ¦ector thet converts the cherecter|st|c ve|ue o¦ en ect|on |nto e representet|ve ve|ue lt ed|usts the ve|ue o¦ the ect|on to eccount ¦or the neture o¦ the ||m|t stete under cons|deret|on end the |o|nt pro|e||||ty o¦ the ect|ons occurr|ng s|mu|teneous|y lt cen essume the ve|ue o¦ !0 ¦or e permenent ect|on or c 0 or c ! or c ? ¦or e ver|e||e ect|on ¹e||e ?!/ shows how cherecter|st|c ve|ues o¦ ver|e||e ect|ons ere converted |nto representet|ve ve|ues ¹h|s te||e |s der|ved ¦rom bS lN !990 |!0| end |ts Net|one| Annex |!0e| / | cherecter|st|c ve|ue o¦ en ect|on es deuned |n Sect|ons ?? end ?3 1ab|o 2.16 Þartia| factors {g | } for uso in vorification of |imit statos in porsistont and transiont dosign situations Limit stato Þormanont actions {@ k } Loading variab|o action {J k,1 } Accompanying variab|o actions {J k,i } d a} £qui|ibrium {£QU} !!0 (09) a !S0 (00) a c 0,| !S0 (00) a b} Strongth at ULS {S1kIC£O} not invo|ving gootochnica| actions Either lxp (6!0) !3S (!0) a !S c 0 !S or the worst case of lxp (6!0e) !3S (!0) a c 0 !S c 0 !S and lxp (6!0|) !?S (!0) a !S c 0 !S c} Strongth at ULS {S1kIC£O} with gootochnica| actions Worst case of Set b !3S (!0) a !S (00) a and Set C !0 !3 d} Sorvicoabi|ity Cherecter|st|c !00 !00 c 0,| !00 lrequent !00 c !,! !00 c ?,| !00 Ques|-permenent !00 c ?,! !00 c ?,| !00 o} Accidonta| dosign situations lxp (6!!e) !0 / d b c !,| (me|n) c ?,| (others) f} Soismic lxp (6!?e/|) !0 / ld c c ?,| koy a Ve|ue |¦ ¦evoure||e (shown |n |rec|ets) b leed|ng ecc|dente| ect|on, / d , |s un¦ectored c Se|sm|c ect|on, / ld d ke¦er to bS lN !990 A!?? 8 NA Notos 1 ¹he ve|ues o¦ c ere g|ven |n ¹e||e ?!/ 2 Ceotechn|ce| ect|ons g|ven |n the te||e ere |esed on Les|gn Approech ! |n C|euse A!3!(S) o¦ bS lN !990, wh|ch |s recommended |n |ts Net|one| Annex EC0: Tables A1.2(A), A1.2(B), A1.2(C), A1.4 & NA 2.9 2.9.1 This publication is available for purchase from www.concretecentre.com 22 Design values at ULS EC0: 6.4.3.2(3) lor the 0lS o¦ strength (S¹k), the des|gner mey choose |etween us|ng lxpress|on (6!0) or the worst cese o¦ lxpress|on (6!0e) or lxpress|on (6!0|) Single variable action At 0lS, the des|gn ve|ue o¦ ect|ons |s e|ther lxp (6!0) !3S O | ÷ !S Ç |,! or the worst cese o¦ lxp (6!0e) !3S O | ÷ c 0,! !S Ç |,! end lxp (6!0|) !?S O | ÷ !S Ç |,! where O | permenent ect|on Ç |,! s|ng|e ver|e||e ect|on c 0,! com||net|on ¦ector ¦or e s|ng|e ver|e||e |oed (see ¹e||e ?!/) EC1-1-1: 3.3.2 1ab|o 2.17 Va|uos of c factors Action c 0 c 1 c 2 |mposod |oads in bui|dings Category A: domestic, residential areas 0/ 0S 03 Category B: office areas 0/ 0S 03 Category C: congregation areas 0/ 0/ 06 Category D: shopping areas 0/ 0/ 06 Category E: storage areas !0 09 0S Category F: traffic area (vehicle weight < 30 kN) 0/ 0/ 06 Category G: traffic area (30 kN < vehicle weight < 160 kN) 0/ 0S 03 Category H: roofs a 0/ 00 00 Snow loads where altitude < 1000 m a.m.s.l. a 0S 0? 00 Wind loads a 0S 0? 00 Temperature effects (non-fire) a 06 0S 00 koy a On roo¦s, |mposed |oeds, snow |oeds end w|nd |oeds shou|d not |e epp||ed together Notos 1 ¹he numer|ce| ve|ues g|ven e|ove ere |n eccordence w|th bS lN !990 end |ts 0l Net|one| Annex 2 Cetegor|es l end l ere essumed to |e es ¦or Cetegory l lxpress|on (6!0) |eeds to the use o¦ g l g C !3S ¦or permenent ect|ons end g l g Q !S0 ¦or ver|e||e ect|ons (g C ¦or permenent ect|ons |s |ntended to |e constent ecross e|| spens) lxpress|on (6!0) |s e|weys eque| to or more conservet|ve then the |ess ¦evoure||e o¦ lxpress|ons (6!0e) end (6!0|) lxpress|on (6!0|) w||| norme||y epp|y when the permenent ect|ons ere not greeter then 4S t|mes the ver|e||e ect|ons (except ¦or storege |oeds, cetegory l |n ¹e||e ?!/, where lxpress|on (6!0e) e|weys epp||es) ¹here¦ore, except |n the cese o¦ concrete structures support|ng storege |oeds where c 0 !0, or ¦or m|xed use, lxpress|on (6!0|) w||| usue||y epp|y ¹hus, ¦or mem|ers support|ng vert|ce| ect|ons et 0lS, !?SO | ÷ !SÇ | w||| |e eppropr|ete ¦or most s|tuet|ons end epp||ce||e to most concrete structures (see l|gure ?S) Compered w|th the use o¦ lxpress|on (6!0), the use o¦ e|ther lxpress|on (6!0e) or (6!0|) |eeds to e more cons|stent re||e||||ty |ndex ecross ||ghtwe|ght end heevywe|ght meter|e|s 2.9.2 EC0: A1.2.2 & NA This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 23 50 40 g k k N / m ( o r k N / m 2 ) q k kN/m (or kN/m 2 ) Use Exp. (6.10b) Use Exp. (6.10a) 30 20 10 0 1 2 3 4 5 6 7 8 9 10 Noto Assum|ng c 0 0/ |e epp||ce||e to e|| erees except storege |iguro 2.5 whon to uso £xp. {6.10a} or £xp. {6.10b} Accompanying variable actions Age|n the des|gner mey choose |etween us|ng lxpress|on (6!0) or the |ess ¦evoure||e o¦ lxpress|ons (6!0e) or (6!0|) EC0: 6.4.3.2(3) l|ther lxp (6!0) !3S O | ÷ !S Ç |,! ÷ S(c 0,| !S Ç |,| ) or the worst cese o¦ lxp (6!0e) !3S O | ÷ c 0,! !S Ç |,! ÷ S(c 0,| !S Ç |,| ) end lxp (6!0|) !?S O | ÷ !S Ç |,! ÷ S(c 0,| !S Ç |,| ) where O | permenent ect|on Ç |,! !st ver|e||e ect|on Ç |,| | th ver|e||e ect|on c 0,! cherecter|st|c com||net|on ¦ector ¦or !st ver|e||e |oed (see ¹e||e ?!/) c 0,| cherecter|st|c com||net|on ¦ector ¦or | th ver|e||e |oed (see ¹e||e ?!/) ln the e|ove, Ç |,! (end c 0,| ) re¦ers to the |eed|ng ver|e||e ect|on end Ç |,| (end c 0,| ) re¦ers to eccompeny|ng |ndependent ver|e||e ect|ons ln genere|, the d|st|nct|on |etween the two types o¦ ect|ons w||| |e o|v|ous (see l|gure ?6), where |t |s not, eech |oed shou|d |n turn |e treeted es the |eed|ng ect|on A|so, the numer|ce| ve|ues ¦or pert|e| ¦ectors g|ven |n the 0l Net|one| Annex |!0e| ere used |n the equet|ons e|ove ¹he ve|ue o¦ c 0 depends on the use o¦ the |u||d|ng end shou|d |e o|te|ned ¦rom the 0l Net|one| Annex ¦or bS lN !990 (see ¹e||e ?!/) EC0: A1.2.2, A1.3.1 & NA q k2 g k2 q k1 g k1 q k1 g k1 q k1 g k1 q k1 q k3 = w k A B C g k1 Noto Cenere||y the ver|e||e ect|ons on e typ|ce| o¦¦|ce ||oc| wou|d |e cons|dered es |e|ng three sets o¦ |ndependent ver|e||e ect|ons lmposed o¦¦|ce |oeds ! on the o¦¦|ce ¦|oors koo¦ |mposed |oed ? v|nd |oed 3 |iguro 2.6 |ndopondont variab|o actions This publication is available for purchase from www.concretecentre.com 24 ¹he express|ons te|e |nto eccount the pro|e||||ty o¦ |o|nt occurrence o¦ |oeds |y epp|y|ng the c 0,| ¦ector to the eccompeny|ng ver|e||e ect|on ¹he pro|e||||ty thet these com||ned ect|ons w||| |e exceeded |s deemed to |e s|m||er to the pro|e||||ty o¦ e s|ng|e ect|on |e|ng exceeded l¦ the two |ndependent ver|e||e ect|ons Ç |,! end Ç |,? ere essoc|eted w|th d|¦¦erent spens end the use o¦ lxpress|on (6!0|) |s eppropr|ete, then |n one set o¦ ene|yses epp|y !?SO | ÷ !SÇ |,! to the Ç |,! spens end !?SO | ÷ c 0| !SÇ |,! to the Ç |,? spens ln essoc|eted ene|yses epp|y !?SO | ÷ c 0,| !SÇ |,! to the Ç |,! spens end !?SO | ÷ !?SÇ |,? to the Ç |,? spens See lxemp|e ?!!? (two ver|e||e ect|ons) Design values at SLS EC0: 6.5 & Table A1.4 ¹here ere three com||net|ons o¦ ect|ons et SlS (or |oed com||net|on et SlS) ¹hese ere g|ven |n ¹e||e ?!S ¹he com||net|on end ve|ue to |e used depends on the neture o¦ the ||m|t stete |e|ng chec|ed Ques|-permenent com||net|ons ere essoc|eted w|th de¦ormet|on, crec| w|dths end crec| contro| lrequent com||net|ons mey |e used to determ|ne whether e sect|on |s crec|ed or not ¹he numer|c ve|ues o¦ c 0 , c ! end c ? ere g|ven |n ¹e||e ?!/ Co||oqu|e||y c 0 hes |ecome |nown es the 'cherecter|st|c' ve|ue c ! hes |ecome |nown es the '¦requent' ve|ue c ? hes |ecome |nown es the 'ques|-permenent' ve|ue EC0: Table A1.4 1ab|o 2.18 Þartia| factors to bo app|iod in tho vorification of tho SLS Combination Permanent actions G k Variable actions Q k Unfavourable a Favourable a Leading b Others b Characteristic O |,sup O |,|n¦ Ç |,! c 0,| Ç |,| Frequent O |,sup O |,|n¦ c !,! Ç |,! c ?,| Ç |,| Quasi-permanent O |,sup O |,|n¦ c ?,! Ç |,! c ?,| Ç |,| koy a Cenere||y O |,sup end O |,|n¦ mey |e te|en es O | See Sect|on ?9S b c ¦ectors ere g|ven |n ¹e||e ?!/ Design values for other limit states loed com||net|ons ere g|ven |n ¹e||e ?!6 ¦or e) lqu||||r|um (lQ0), |) Strength et 0lS not |nvo|v|ng geotechn|ce| ect|ons, c) Strength et 0lS w|th geotechn|ce| ect|ons, d) Serv|cee||||ty, e) Acc|dente| end ¦) Se|sm|c des|gn s|tuet|ons 2.9.3 2.9.4 This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 25 Variations in permanent actions vhen the ver|et|on o¦ e permenent ect|on |s not sme|| then the upper (O ||,sup ) end the |ower (O ||,|n¦ ) cherecter|st|c ve|ues (the 9S% end S% ¦rect||e ve|ues respect|ve|y) shou|d |e este|||shed ¹h|s procedure |s necessery on|y when the coe¦uc|ent o¦ ver|et|on ( !00 t stenderd dev|et|on/ meen) |s greeter then !0 ln terms o¦ permenent ect|ons, ver|et|ons |n the se|¦-we|ght o¦ concrete |n concrete ¦remes ere cons|dered sme|| EC0: 4.1.2, 4.1.2 (3) PD 6687 [6] : 2.8.4 At 0lS where the ver|et|on |s not sme||, g C|,sup shou|d |e used w|th O ||,sup end g C|,|n¦ w|th O ||,|n¦ S|m||er|y, where the ver|et|on |s not sme||, et SlS O ||,sup shou|d |e used where ect|ons ere un¦evoure||e end O ||,|n¦ used where ¦evoure||e vhere chec|s, note||y chec|s on stet|c equ||||r|um (lQ0), ere very sens|t|ve to ver|et|on o¦ the megn|tude o¦ e permenent ect|on ¦rom one p|ece to enother, the ¦evoure||e end un¦evoure||e perts o¦ th|s ect|on shou|d |e cons|dered es |nd|v|due| ect|ons ln such 'very sens|t|ve' ver|ucet|ons g C,sup end g C,|n¦ shou|d |e used EC0: 6.4.3 (4) Load arrangements of actions: introduction ¹he process o¦ des|gn|ng concrete structures |nvo|ves |dent|¦y|ng re|event des|gn s|tuet|ons end ||m|t stetes ¹hese |nc|ude pers|stent, trens|ent or ecc|dente| s|tuet|ons ln eech des|gn s|tuet|on the structure shou|d |e ver|ued et the re|event ||m|t stetes EC0: 3.2 ln the ene|ys|s o¦ the structure et the ||m|t stete |e|ng cons|dered, the mex|mum e¦¦ect o¦ ect|ons shou|d |e o|te|ned us|ng e ree||st|c errengement o¦ |oeds Cenere||y ver|e||e ect|ons shou|d |e errenged to produce the most un¦evoure||e e¦¦ect, ¦or exemp|e to produce mex|mum overturn|ng moments |n spens or mex|mum |end|ng moments |n supports lor |u||d|ng structures, des|gn concentretes me|n|y on the 0lS, the u|t|mete ||m|t stete o¦ strength (S¹k), end SlS, the serv|cee||||ty ||m|t stete lowever, |t |s essent|e| thet e|| ||m|t stetes ere cons|dered ¹he ||m|t stetes o¦ equ||||r|um (lQ0), strength et 0lS w|th geotechn|ce| ect|ons (S¹k/ClO) end ecc|dente| s|tuet|ons must |e te|en |nto eccount es eppropr|ete EC0: 3.3, 3.4, 6.4, 6.5 Load arrangements according to the UK National Annex to Eurocode ln |u||d|ng structures, eny o¦ the ¦o||ow|ng sets o¦ s|mp||ued |oed errengements mey |e used et 0lS end SlS (See l|gure ?/) Cl. 5.1.3 & NA ¹he more cr|t|ce| o¦ N e) e|ternete spens cerry|ng g C O | ÷ g Q Ç | w|th other spens |oeded w|th g C O | , end |) eny two ed|ecent spens cerry|ng g C O | ÷ g Q Ç | w|th other spens |oeded w|th g C O | Or the more cr|t|ce| o¦ N e) e|ternete spens cerry|ng g C O | ÷ g Q Ç | , w|th other spens |oeded w|th g C O | , end |) e|| spens cerry|ng g C O | ÷ g Q Ç | Or, ¦or s|e|s on|y, e|| spens cerry|ng N g C O | ÷ g C O | , prov|ded the ¦o||ow|ng cond|t|ons ere met ln e one-wey spenn|ng s|e| the eree o¦ eech |ey exceeds 30 m L ? (e |ey |s de¦|ned es e str|p ecross the ¦u|| w|dth o¦ e structure |ounded on the other s|des |y ||nes o¦ support) ¹he ret|o o¦ the ver|e||e ect|on, L Ç | , to the permenent ect|on, O | , does not exceed !?S ¹he megn|tude o¦ the ver|e||e ect|on exc|ud|ng pert|t|ons does not exceed S |N/m L ? vhere ene|ys|s |s cerr|ed out ¦or the s|ng|e |oed cese o¦ e|| spens |oeded, the resu|t|ng moments, except those et cent||evers, shou|d |e reduced |y ?0%, w|th e consequent|e| |ncreese |n the spen moments 2.9.5 2.10 2.11 This publication is available for purchase from www.concretecentre.com 26 g Q Ç | g C O | a) Alternate spans loaded b) Adjacent spans loaded c) All spans loaded g Q Ç | g Q Ç | g Q Ç | g C O | g C O | Noto vh||st the use o¦ lxp (6!0) |s |nd|ceted, these errengements mey eque||y |e used w|th lxp (6!0e) or (6!0|) |iguro 2.7 Load arrangomonts for boams and s|abs according to Uk NA to £urocodo This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 27 2.12.1 Continuous beam in a domestic structure Determine the appropriate load combination and ultimate load for a continuous beam of four 6 m spans in a domestic structure supporting a 175 mm slab at 6 m centres. 6000 mm 6000 mm 6000 mm 6000 mm q k g k A B C D E Figure 2.8 Continuous beam in a domestic structure a) Actions kN/m Permanent action, g k Self-weight, 175 mm thick slabs : 0.17 x 25 x 6.0 = 26.3 E/o self-weight downstand 800 × 225 : 0.80 x 0.225 x 25 = 4.5 50 mm screed @ 22 kN/m 3 : 0.05 x 22 x 6.0 = 6.6 Finishes and services : 0.50 x 6.0 = 3.0 Dividing wall 2.40 × 4.42 (200 mm dense blockwork with plaster both sides) = 10.6 Total g k = 51.0 Variable action, q k Imposed, dwelling @ 1.5 kN/m 2 : 1.5 x 6.0 = 9.0 Total q k = 9.0 Ultimate load, n Assuming use of Exp. (6.10), n = 1.35 × 51 + 1.5 × 9.0 = = 82.4 Assuming use of worst case of Exp. (6.10a) or Exp. (6.10b) Exp. (6.10a): n = 1.35 × 51 + 0.7 × 1.5 × 9.0 = = 78.3 Exp. (6.10b): n = 1.25 × 51 + 1.5 × 9.0 = = 77.3 In this case Exp. (6.10a) would be critical ‡ =ultimate load = 78.3 ‡ This could also be determined from Figure 2.5 or by determining that g k > 4.5q k lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Continuous beam in a domestic structure 2.12 Examples of loading This publication is available for purchase from www.concretecentre.com 28 2.12.2 Continuous beam in mixed use structure Determine the worst case arrangements of actions for ULS design of a continuous beam supporting a 175 mm slab @ 6 m centres. Note that the variable actions are from two sources as defined in Figure 2.9.: 6000 mm 6000 mm 6000 mm 6000 mm Office use @ 2.5 kN/m 2 Shopping use @ 4.0 kN/m 2 c 0 = 0.7 c 0 = 0.7 q k1 = 15 kN/m A B C D E q k2 = 24 kN/m g k = 51 kN/m Figure 2.9 Continuous beam in mixed-use structure a) Load combination Load combination Exp. (6.10a) or Exp. (6.10b) will be used, as either will produce a smaller total load than Exp. (6.10). It is necessary to decide which expression governs. i) Actions kN/m Permanent action As before, Example 2.12.1 g k = 51.0 Variable action Office @ 2.5 kN/m 2 q k1 = 15.0 Shopping @ 4.0 kN/m 2 q k2 = 24.0 Ultimate load, n For office use: Exp. (6.10a): n = 1.35 × 51 + 0.7 × 1.5 × 15.0 = 84.6 Exp. (6.10b): n = 1.25 × 51 + 1.5 × 15.0 = 86.3 For shopping use: Exp. (6.10a): n = 1.35 × 51 + 1.5 × 0.7 × 24.0 = 94.1 Exp. (6.10b): n = 1.25 × 51 + 1.5 × 24.0 = 99.8 By inspection Exp. (6.10b) governs in both cases ‡ b) Arrangement of ultimate loads As the variable actions arise from different sources, one is a leading variable action and the other is an accompanying variable action. The unit loads to be used in the various arrangements are: ‡ This could also be determined from Figure 2.5 or by determining that g k > 4.5q k EC1-1-1: 6.3.1.1 & NA, EC0: A.1.2.2. & NA lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Continuous beam in mixed use structure This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 29 i) Actions kN/m Permanent 1.25 × 51.0 = 63.8 Variable Office use as leading action, g Q Q k = 1.5 × 15 = 22.5 as accompanying action, c 0 g Q Q k = 0.7 × 1.5 × 15 = 15.75 Shopping use as leading action, g Q Q k = 1.5 × 24 = 36.0 as accompanying action, c 0 g Q Q k = 0.7 × 1.5 × 24 = 25.2 ii) For maximum bending moment in span AB The arrangement and magnitude of actions of loads are shown in Figure 2.10. The variable load in span AB assumes the value as leading action and that in span CD takes the value as an accompanying action. A B C D E Leading variable action g Q q k1 = 22.5 kN/m Accompanying variable action c Q g Q q k2 = 25.2 kN/m Permanent action g G g k = 63.8 kN/m Figure 2.10 For maximum bending moment in span AB p iii) For maximum bending moment in span CD The load arrangement is similar to that in Figure 2.10, but now the variable load in span AB takes its value as an accompanying action (i.e. 15.75 kN/m) and that in span CD assumes the value as leading action (36 kN/m). A B C D E Accompanying variable action c Q g Q q k1 = 15.8 kN/m Leading variable action g Q q k2 = 36.0 kN/m Permanent action g G,inf g k = 63.8 kN/m Figure 2.11 For maximum bending moment in span CD p This publication is available for purchase from www.concretecentre.com 30 iv) For maximum bending moment at support B The arrangement of loads is shown in Figure 2.12. As both spans AB and BC receive load from the same source, no reduction is possible (other than that for large area ( g (other than that for large area ‡ ). )). EC1-1-1: 6.3.1.1 (10) & NA & NA A B C D E Leading variable action g Q gg q k1 = 22.5 kN/m Permanent action g G gg g k = 63.8 kN/m Figure 2.12 For maximum bending moment at support B pp v) For maximum bending moment at support D The relevant arrangement of loads is shown in Figure 2.13. Comments made in d) also apply here. A B C D E Leading variable action g Q gg q k2 = 36 kN/m Permanent action g G gg g k = 63.8 kN/m Figure 2.13 For maximum bending moment at support D pp vi) For critical curtailment and hogging in span CD The relevant arrangement of loads is shown in Figure 2.14. A B C D E Leading variable action g Q q k2 = 36.0 kN/m Accompanying variable action p c 0gQ q k1 = 15.8 kN/m Pe n rmanent actionn g G,in gg f g k m = 51 kN/mm Figure 2.14 For curtailment and hogging in span CD p Eurocode 2 requires that all spans should be loaded with either g G,sup gg or g G,inf gg (as f per Table 2.16). As illustrated in Figure 2.14, using g G,inf gg = 1.0 might be critical for f curtailment and hogging in spans. curtailment and hogging in spans curtailment and hogging in spans. Cl. 2.4.3(2) ‡ Variable actions may be subjected to reduction factors: a A , according to the pp ( area supported (m area supported (m 2 ), ), ), aa AAA = 1.0 – A/1000 1.0 A/1000 ≥≥ 0.75. 0.75. EC1-1-1: 6.3.1.2 (10) & NA & NA This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 31 lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Propped cantilever 2.12.3 Propped cantilever Determine the Equilibrium, ULS and SLS (deformation) load combinations for the propped cantilever shown in Figure 2.15. The action P at the end of the cantilever arises from the permanent action of a wall. q k g k A B P C Figure 2.15 Propped cantilever beam and loading pp For the purposes of this example, the permanent action P is considered to be from a separate source than the self-weight of the structure so both g G,sup and g G,inf need to be considered. a) Equilibrium limit state (EQU) for maximum uplift at A 0.0q k = 0 1.5q k g Ginf g k = 0.9g k g Gk,sup g k = 1.1g k g Gk,sup P= 1.1P A B C p EC0: Table 1.2(B), Note 3 EC0: Table A1.2 (A) & NA EC0: 6.4.3.1 (4), Table A1.2 (A) & NA b) Ultimate limit state (ULS) i) For maximum moment at B and anchorage of top reinforcement BA g Gk,sup g k = 1.35g k g Gk,sup P= 1.35P g Q q k = 1.5q k A B C Figure 2.17 ULS: maximum moment at B Notes g Gk,inf g k = 1.0 g k may be critical in terms of curtailment of top bars BA. EC0: Tables A1.1, A1.2 (B) & NA This publication is available for purchase from www.concretecentre.com 32 ii) For maximum sagging moment AB g Q gg q k = 1.5q k g Gk,sup gg g k = 1.35g k g Gk,sup gg P= PP 1.1P A B C Figure 2.18 ULS: maximum span moment AB p Notes Notes 1 Depending on the magnitude of g k , q k length AB and BC, g Gk,inf gg g k (= 1.0 g k ) may be more critical for span moment. 2 The magnitude of the load combination indicated are those for Exp. (6.10) of BS EN 1990. The worst case of Exp. (6.10a) and Exp. (6.10b) may also have been used. 3 Presuming supports A and B were columns then the critical load combination for Column A would be as Figure 2.18. For column B the critical load combination might be either as Figure 2.17 or 2.18. EC0: Table A1.1, A1.2 (B) & NA c) Serviceability limit state (SLS) of deformation: (quasi-permanent loads) i) For maximum deformation at C 1.0g k 1.0P 1.0c 2 cc q k =0.3*q k *Assuming office area A B C Figure 2.19 SLS: maximum deformation at C EC0: Tables A1.1, A1.2.2, A1.4 & NA ii) For maximum deformation AB 1.0P A B C * Assuming office area 1.0c 2 cc q k =0.3*q k 1.0g k Figure 2.20 SLS: maximum deformation AB Notes Notes Quasi-permanent load combinations may also be used for calculations of crack widths or controlling cracking, i.e. the same load combinations as shown in Figures 2.19 and 2.20 may be used to determine SLS moment to determine stress in reinforcement. The characteristic and/or frequent combinations may be appropriate for other SLS limit states: for example, it is recommended that the frequent combination is used to determine whether a member has cracked or not. This publication is available for purchase from www.concretecentre.com Ane|ys|s, ect|ons end |oed errengements 33 lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP-041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Overall stability 2.12.4 Overall stability (EQU) For the frame shown in Figure 2.21, identify the various load arrangements to check overall stability (EQU) against overturning. Assume that the structure is an office block and that the loads q k2 and q k3 may be treated as arising from one source. w k g k1 q k1 g k2 q k2 g k3 q k3 A B Figure 2.21 Frame configuration a) EQU – Treating the floor imposed load as the leading variable r action Permanent action, PP favourable 0.9g k1 Permanent action, PP favourable 0.9g k2 Permanent action, PP favourable 0.9g k3 Accompanying variable action = g Qk gg c o cc w k ww = 1.5 x 0.5 x w k ww = 0.75 w k ww Accompanying va n = riable actioon on g Qk gg c o cc q k1 = 1.05q k1 Permanent action, unfavour ble = aab a g Gk,sup gg g k1 = 1.1g k1 Lead variable action = g Qk gg q 2 kk2 k2 = 1.5q k2 Permanent action, unfavour ble = aab a g Gk,sup gg g k2 = 1.1g k2 Lead variable action = g Qk gg q 3 kk3 k3 = 1.5q k3 Permanent action, unfavour ble = aab a g Gk,sup gg g k3 = 1.1g k3 A B Figure 2.22 Frame with floor variable action as leading variable action Tables 2.16 Tables 2.16 & 2.17 & 2.17 See Table 2.17 for values of c 0 cc This publication is available for purchase from www.concretecentre.com 34 b) EQU – Treating the roof imposed load as the leading variable f action 0.9g k1 0.9g k2 0.9g k3 0.75 w k ww 1.5q k1 1.1g k1 1.5 x 0.7 x q k2 = 1.05 q k2 1.1g k2 1.5 x 0.7 x q k3 = 1.05 q k3 1.1g k3 A B tion Figure 2.23 Frame with roof variable action as leading variable actt Tables 2.16 Tables 2.16 & 2.17 & 2.17 c) EQU – Treating wind as the leading variable action 0.9g k1 0.9g k2 0.9g k3 1.5 w k ww 1.5 x 0.7 x q k1 = 1.05 q k1 1.1g k1 1.5 x 0.7 x q k2 = 1.05 q k2 1.1g k2 1.5 x 0.7 x q k3 = 1.05 q k3 1.1g k3 A B Figure 2.24 Frame with wind as lead variable action Tables 2.16 Tables 2.16 & 2.17 & 2.17 This publication is available for purchase from www.concretecentre.com 35 3 S|e|s Slabs General ¹he ce|cu|et|ons |n th|s sect|on ere presented |n the ¦o||ow|ng su|-sect|ons 3! A s|mp|y supported one-wey s|e| 3? A cont|nuous one-wey s|e| 33 A cont|nuous r|||ed s|e| 34 A |ey o¦ e ¦et s|e| 3S A ste|r ¦|ght ¹hese ce|cu|et|ons ere |ntended to show whet m|ght |e deemed typ|ce| hend ce|cu|et|ons ¹hey ere |||ustret|ve o¦ the Code end ere not necesser||y |est prect|ce ¹he urst three su|-sect|ons |nc|ude dete|||ng chec|s eg curte||ment |engths determ|ned str|ct|y |n eccordence w|th the prov|s|ons o¦ bS lN !99?÷!÷! ¹he ¦et s|e| ce|cu|et|on |s supp|emented |y e commentery A genere| method o¦ des|gn|ng s|e|s |s shown |e|ow Leterm|ne des|gn ||¦e N EC0 & NA Table NA.2.1 Assess ect|ons on the s|e| N EC1 & NA Assess dure||||ty requ|rements end determ|ne N concrete strength Table 4.1 BS 8500-1: Tables A4 & A5 Chec| cover requ|rements ¦or eppropr|ete ¦|re N res|stence per|od EC2-1-2: Tables 5.8, 5.9, 5.10 & 5.11 Ce|cu|ete m|n|mum cover ¦or dure||||ty, ¦|re end N |ond requ|rements Cl. 4.4.1 Leterm|ne wh|ch com||net|ons o¦ ect|ons epp|y N EC0 & NA Tables NA.A1.1 & NA.A1.2 (B) Leterm|ne |oed|ng errengements N Cl. 5.1.3(1) & NA Ane|yse structure to o|te|n cr|t|ce| moments end sheer N ¦orces Cl. 5.4, 5.5, 5.6 Les|gn ¦|exure| re|n¦orcement N Cl. 6.1 Chec| de¦|ect|on N Cl. 7.4 Chec| sheer cepec|ty N Cl. 6.2 Other des|gn chec|s N Chec| m|n|mum re|n¦orcement Chec| crec||ng (s|.e or spec|ng o¦ |ers) Chec| e¦¦ects o¦ pert|e| ux|ty Chec| secondery re|n¦orcement Cl. 9.3.1.1(1), 9.2.1.1(1) Cl. 7.3, Tables 7.2N & 7.3N Cl. 9.3.1.2(2) Cl. 9.3.1.1(2), 9.3.1.4(1) Chec| curte||ment N Cl. 9.3.1.1(4), 9.2.1.3, Fig. 9.2 Chec| enchorege N Cl. 9.3.1.2, 8.4.4, 9.3.1.1(4) Cl. 9.2.1.5(1), 9.2.1.5(2) Chec| |eps N Cl. 8.7.3 3 3.0 This publication is available for purchase from www.concretecentre.com 36 Simply supported one-way slab ¹h|s ce|cu|et|on |s |ntended to show e typ|ce| |es|c hend ce|cu|et|on 3.1 lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Simply supported one-way slab A 175 mm thick slab is required to support screed, finishes, an office variable action of 2.5 kN/m 2 and demountable partitions (@ 2 kN/m). The slab is supported on load-bearing block walls. f ck = 30 MPa, f yk = 500 MPa. Assume a 50-year design life and a requirement for 1 hour resistance to fire. q k = 3.3 kN/m 2 g k = 5.9 kN/m 2 4800 Figure 3.1 Simply supported one-way slab p pp 3.1.1 Actions kN/m 2 Permanent: Self-weight 0.175 × 25 = 4.4 EC1-1-1: Table A1 50 mm screed = 1.0 Finishes, services = 0.5 Total g k = 5.9 Variable: Offices, general use B1 = 2.5 EC1-1-1: Tables 6.1, 6.2 & NA Movable partitions @ 2.0 kN/m = 0.8 Total ¡ k = 3.3 EC1-1-1: 6.3.12(8) 3.1.2 Cover Nominal cover, c nom : c nom = c min + Dc dev where c min = max[c min,b ; c min,dur ] where c min,b = minimum cover due to bond = diameter of bar Assume 12 mm main bars. c min,dur = minimum cover due to environmental conditions Assuming XCI and using C30/37 concrete, c min,dur = 15 mm Dc dev = allowance in design for deviation. Assuming no measurement of cover, Dc dev = 10 mm = c nom = 15 + 10 = 25 mm Exp. (4.1) Cl. 4.4.1.2(3) Table 4.1. BS 8500-1: Table A4. Cl. 4.4.1.2(3) This publication is available for purchase from www.concretecentre.com 37 37 3! S|mp|y supported one-wey s|e| Fire: Check adequacy of section for 1 hour fire resistance (i.e. REI 60). Thickness, h s,min = 80 mm cf. 175 mm proposed = OK Axis distance, a min = 20 mm cf. 25 + f/2 = 31 i.e. not critical = OK = choose c nom = 25 mm EC2-1-2: 4.1(1), 5.1(1) & Table 5.8 3.1.3 Load combination (and arrangement) Ultimate load, n: By inspection, BS EN 1990 Exp. (6.10b) governs = n = 1.25 × 5.9 + 1.5 × 3.3 = 12.3 kN/m 2 3.1.4 Analysis Design moment: M Ed = 12.3 × 4.8 2 /8 = 35.4 kNm Shear force: V = 12.3 × 4.8/2 = 29.5 kN/m Fig. 2.5 ECO: Exp. (6.10b) 3.1.5 Flexural design Effective depth: d = 175 − 25 − 12/2 = 144 mm Flexure in span: K = M Ed /bd 2 f ck = 35.4 × 10 6 /(1000 × 144 2 × 30) = 0.057 z/d = 0.95 z = 0.95 × 144 = 137 mm A s = M Ed /f yd z = 35.4 × 10 6 /(137 × 500/1.15) = 594 mm 2 /m (r = 0.41%) Try H12 @ 175 B1 (645 mm 2 /m) Fig. 3.5 Appendix A1 Table C5 3.1.6 Deflection Check span-to-effective-depth ratio. Basic span-to-effective-depth ratio for r = 0.41% = 20 A s,prov /A s,req = 645/599 = 1.08 Max. span = 20 × 1.08 × 144 = 3110 mm i.e. < 4800 mm = no good Consider in more detail: Allowable l/d = N × K × F1 × F2 × F3 where N = 25.6 (r = 0.41%, f ck = 30 MPa) K = 1.0 (simply supported) F1 = 1.0 (b eff /b w = 1.0) F2 = 1.0 (span < 7.0 m) F3 = 310/ s s < 1.5 Appendix B Table 7.4N & NA Exp. (7.17) Cl. 7.4.2, Appendix C7, Tables C10-C13 This publication is available for purchase from www.concretecentre.com 38 38 where ‡ s s s = s su s (A s,req / q A s,prov ) 1/d where s su s { 242 MPa (From Figure C3 and g k /q k = 1.79, c 2 cc = 0.3, g G gg = 1.25) d = redistribution ratio = 1.0 d = s s s { 242 × 594/645 = 222 = F3 = 310/222 = 1.40 < 1.5 = Allowable l/ ll d = 25.6 × 1.40 = 35.8 d Actual l/ ll d = 4800/144 = 33.3 d = OK Use H12 @ 175 B1 (645 mm 2 /m) Cl. 7.4.2, Exp. (7.17) Table 7.4N, & NA Table NA.5: Note 5 Figure C3 Figure C3 Figure C3 Figure C3 3.1.7 Shear By inspection, OK However, if considered critical: V = 29.5 kN/m as before V V Ed VV = 29.5 – 0.14 × 12.3 = 27.8 kN/m v Ed vv = 27.8 × 10 3 /144 × 10 3 = 0.19 MPa v Rd,c vv = 0.53 MPa = No shear reinforcement required q Cl. 6.2.1(8) Cl. 6.2.2(1); Table C6 Table C6 3.1.8 Summary of design H12 @ 175 Figure 3.2 Simply supported slab: summary p pp 3.1.9 Detailing checks It is presumed that the detailer would take the design summarised above and detail the slab to normal best practice, e.g. to SMDSC [9] or to How to design concrete structures using Eurocode 2, 2 [8] Chapter 10, Detailing. This would usually include dimensioning and detailing curtailment, laps, U-bars and also undertaking the other checks detailed below. See also 3.2.10 detailing checks for a continuous one-way slab. a) Minimum areas Minimum area of reinforcement: A s,min = 0.26 (f ctm f /f yk ff ) b t bb J ~ 0.0013 b t bb d where b t bb = width of tension zone f ctm f = 0.30 × f ck f 0.666 Cl. 9.3.1.1, 9.2.1.1 Table 3.1 ‡ See Appendix B1.5 This publication is available for purchase from www.concretecentre.com S|e|s 39 39 3! S|mp|y supported one-wey s|e| A s,min = 0.26 × 0.30 × 30 0.666 × 1000 × 144/500 = 216 mm 2 /m (r = 0.15%) r = H12 @ 175 B1 OK Crack control: OK by inspection. Maximum spacing of bars: < 3h < 400 mm OK Secondary reinforcement: 20% A s,req = 0.2 × 645 = 129 mm q 2 /m Use H10 @ 350 (224) B2 Edges: effects of assuming partial fixity along edge Top steel required = 0.25 × 594 = 149 mm 2 /m Use H10 @ 350 (224) T2 B2 as U-bars extending 960 mm into slab § Table 7.2N & NA Cl. 9.3.1.1.(3) Cl. 9.3.1.1.(2) Cl. 9.3.1.2.(2) b) Curtailment Curtailment main bars: Curtail main bars 50 mm from or at face of support. At supports: 50% of A s to be anchored from face of support. Use H12 @ 350 B1 T1 U-bars In accordance with SMDSC [9] detail MS3 lap U-bars 500 mm with main steel, curtail T1 leg of U-bar 0.1l (= say 500 mm) from face l of support. SMDSC [9] : Fig. 6.4; How to [8] : Detailing Cl. 9.3.1.2.(1) § A free unsupported edge is required to use ‘longitudinal and transverse reinforcement’ generally using U-bars with legs at least 2h long. For slabs 150 mm deep or greater, SMDSC [9] standard detail recommends U-bars lapping 500 mm with bottom steel and extending 0.1l top into span. l Cl. 9.3.1.4.(1) This publication is available for purchase from www.concretecentre.com 40 40 3.2 Continuous one-way solid slab ¹h|s ce|cu|et|on |s |ntended to show |n dete|| the prov|s|ons o¦ des|gn|ng e s|e| to lurocode ? us|ng essent|e||y the seme s|e| es used |n lxemp|e 3! lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Continuous one-way solid slab A 175 mm thick continuous slab is required to support screed, finishes, an office variable action of 2.5 kN/m 2 and demountable partitions (@ 2 kN/m). The slab is supported on 200 mm wide load-bearing block walls at 6000 mm centres. f ck = 30, f yk = 500 and the design life is 50 years. A fire resistance of 1 hour is required. q k = 3.3 kN/m 2 g k = 5.9 kN/m 2 5800 5800 200 200 Figure 3.3 Continuous solid slab 3.2.1 Actions kN/m 2 Permanent: As Section 3.1.1 g k = 5.9 Variable: As Section 3.1.1 ¡ k = 3.3 EC1-1-1: 6.3.1.2(8) 3.2.2 Cover Nominal cover, c nom : As Section 3.1.2 c nom = 25 mm 3.2.3 Load combination (and arrangement) Fig. 2.5 EC0: Exp. (6.10b) Ultimate action (load): As Section 3.1.3, BS EN 1990 Exp. (6.10b) governs = n = 1.25 × 5.9 + 1.5 × 3.3 = 12.3 kN/m 2 3.2.4 Analysis Clear span, l n = 5800 mm a 1 = min[h/2; t/2] = min[175/2; 200/2] = 87.5 mm Cl. 5.3.2.2(1) a 2 = min[h/2; t/2] = min[175/2; 200/2] = 87.5 mm l eff = 5975 mm This publication is available for purchase from www.concretecentre.com S|e|s 41 41 3? Cont|nuous one-wey so||d s|e| Bending moment: End span M Ed = 0.086 × 12.3 × 5.975 2 = 37.8 kNm/m Cl. 5.1.1(7) Table C2 1st internal support M Ed = 0.086 × 12.3 × 5.975 2 = 37.8 kNm/m Internal spans M Ed = 0.063 × 12.3 × 5.975 2 = 27.7 kNm/m and supports Shear: End support V Ed = 0.40 × 12.3 × 5.975 = 29.4 kN/m 1st interior support V Ed = 0.60 × 12.3 × 5.975 = 44.1 kN/m 3.2.5 Flexural design: span a) End span (and 1st internal support) Effective depth, d: d = h − c nom − f/2 = 175 − 25 − 12/2 = 144 mm Relative flexural stress, K: K = M Ed /bd 2 f ck = 37.8 × 10 6 /1000 × 144 2 × 30 = 0.061 K' = 0.207 or restricting x/d to 0.45 K' = 0.168 = by inspection, section is under-reinforced (i.e. no compression reinforcement required). Appendix A1 Lever arm, z: z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d ‡ = (144/2) [1 + (1 − 3.53 × 0.061) 0.5 ] = 0.945d = 136 mm Fig. 3.5 Appendix A1 Area of steel, A s : A s = M Ed /f yd z = 37.8 × 10 6 /(500/1.15 × 136) = 639 mm 2 /m (r = 0.44%) Try H12 @ 175 B1 (645 mm 2 /m) b) Internal spans and supports Lever arm, z: By inspection, z = 0.95d = 0.95 × 144 = 137 mm Fig. 3.5 Appendix A1 Area of steel, A s : A s = M Ed /f yd z = 27.7 × 10 6 /(500/1.15 × 137) = 465 mm 2 /m (r = 0.32%) Try H12 @ 225 B1 (502 mm 2 /m) ‡ Designers may choose to use another form of this equation: z/d = 0.5 + (0.25 − 0.882K) 0.5 ≤ 0.95 This publication is available for purchase from www.concretecentre.com 42 42 3.2.6 Deflection: end span Check end span-to-effective-depth ratio. Allowable l/d = d N × N K × F1 × F2 × F3 Appendix B Appendix B where N = basic effective depth to span ratio: N r = 0.44% r r 0 = f ck f 0.5 × 10 −3 = 0.55% = use Exp. (7.16a) Cl. 7.4.2(2) N = 11 + 1.5 N f ck f 0.5 r 0 /r + 3.2 r f ck f 0.5 (r 0 /r − 1) r 1.5 = 11 + 1.5 × 30 0.5 × 0.55/0.44 + 3.2 × 30 0.5 (0.55/0.44 – 1) 1.5 = 11.0 + 10.3 + 2.2 = 23.5 Exp. (7.16a) K = structural system factor = 1.3 (end span of continuous slab) Cl. 7.4.2 F1 = flanged section factor = 1.0 (b eff /b w = 1.0) w Cl. 7.4.2 F2 = factor for long spans associated with brittle partitions = 1.0 (span < 7.0 m) Cl. 7.4.2 F3 = 310/ s s s ≤ 1.5 Cl. 7.4.2, Exp. (7.17) Table 7.4N & NA, Table NA.5: Note 5 where ‡ s s s = (f yk f / g S gg ) (A s,req / q A s,prov ) (SLS loads/ULS loads (1/d) = f yd ff × (A s,req / q A s,prov ) × (g k + c 2 cc q k )/(g G gg g k + g Q gg q k ) (1/d) = (500/1.15) × (639/645) × [(5.9 + 0.3 × 3.3)/12.3] × 1.08 § = 434.8 × 0.99 × 0.56 × 1.08 = 260 MPa Exp. (7.17) EC0: A1.2.2 Table C14 Table C14 F3 = 310/260 = 1.19 Note: A s,prov /A s,req ≤ 1.50 Table 7.4N & NA, Table NA.5: Note 5 Allowable l/d = d N × N K × F1 × F2 × F3 = 23.5 × 1.3 × 1.0 × 1.19 = 36.4 Max. span = 36.4 × 144 = 5675 mm, i.e. < 5795 mm = No good Try increasing reinforcement to H12 @ 150 B1 (754 mm 2 /m) s s s = 434.8 × 639/754 × 0.56 × 1.08 = 223 F3 = 310/223 = 1.39 Allowable l/d = 23.5 × 1.3 × 1.0 × 1.39 d = 42.5 ‡ See Appendix B1.5 § The use of Table C3 implies certain amounts of redistribution, which are defined in Table C14. This publication is available for purchase from www.concretecentre.com S|e|s 43 43 3? Cont|nuous one-wey so||d s|e| Max. span = 42.5 × 144 = 6120 mm, i.e. > 5795 mm OK = H12 @ 150 B1 (754 mm 2 /m) OK 3.2.7 Deflection: internal span Check internal span-to-effective-depth ratio. Allowable l/d = N × K × F1 × F2 × F3 where N = basic effective depth to span ratio: N r = 0.32% r r 0 = f ck f 0.5 × 10 −3 = 0.55% = use Exp. (7.16a) Cl. 7.4.2(2) N = 11 + 1.5 N f ck f 0.5 r 0 /r + 3.2 r f ck f 0.5 (r 0 /r − 1) r 1.5 = 11 + 1.5 × 30 0.5 × 0.55/0.32 + 3.2 × 30 0.5 (0.55/0.32 – 1) 1.5 = 11.0 + 14.1 + 10.7 = 35.8 Exp. (7.16a) K = structural system factor = 1.5 (interior span of continuous slab) Cl. 7.4.2 F1 = flanged section factor = 1.0 (b eff /b w bb = 1.0) w Cl. 7.4.2 F2 = factor for long spans associated with brittle partitions = 1.0 (span < 7.0 m) Cl. 7.4.2 F3 = 310/ s s s ≤ 1.5 Cl. 7.4.2, Exp. (7.17), Table 7.4N & NA, Table NA.5 Note 5. where s s s = f yd ff × (A s,req / q A s,prov ) × (g k + c 2 cc q k )/(g G gg g k + g Q gg q k ) (1/d) = (500/1.15) × (465/502) × [(5.9 + 0.3 × 3.3)/12.3] × 1.03 = 434.8 × 0.93 × 0.56 × 1.03 = 233 MPa Exp. (7.17) EC0: A1.2.2 Table C14 Table C14 F3 = 310/233 = 1.33 Allowable l/d = d N × N K × F1 × F2 × F3 = 35.8 × 1.5 × 1.0 × 1.33 = 71.4 Max. span = 71.4 × 144 = 10280 mm i.e. > 5795 mm OK Use H12 @ 225 B1 (502 mm 2 /m) in internal spans p 3.2.8 Shear Design shear force, V Ed VV : At d from face of end support, d V Ed VV = 29.4 – (0.144 + 0.0875) × 12.3 = 26.6 kN/m Cl. 6.2.1(8) At d from face of 1st interior support, d V Ed VV = 44.1 − (0.144 + 0.0875) × 12.3 = 41.3 kN/m Shear resistance, V Rd,c VV : V Rd,c VV = (0.18/ g C gg )k(100 r l f ck f ) 0.333 b w bb d ≥ 0.0035k 1.5 k f ck f 0.5 b w bb d Cl. 6.2.2(1) This publication is available for purchase from www.concretecentre.com 44 44 where k = 1 + (200/ k d) 0.5 ≤ 2.0 as d < 200 mm d k = 2.0 k r l = A sl /bd Assuming 50% curtailment (at end support) = 50% × 754/(144 × 1000) = 0.26% V Rd,c VV = (0.18/1.5) × 2.0 × (100 × 0.26/100 × 30) 0.33 × 1000 × 144 = 0.12 × 2 × 1.97 × 1000 × 144 = 0.47 × 1000 × 144 = 68.1 kN/m But V Rd,cmin VV = 0.035k 1.5 f ck f 0.5 b w bb d where k = 1 + (200/ k d) 0.5 ≤ 2.0; as before k = 2.0 k V Rd,cmin VV = 0.035 × 2 1.5 × 30 0.5 × 1000 × 144 = 0.54 × 1000 × 144 = 77.6 kN/m = V Rd,c VV = 77.6 kN/m = OK, no shear reinforcement required at end or 1st internal supports = H12 @ 150 B1 & H12 @ 175 T1 OK By inspection, shear at other internal supports OK. 3.2.9 Summary of design f ck f = 30 MPa PP c nom = 25 mm H12 @ 150 H12 @ 225 H12 @ 175 Figure 3.4 Continuous solid slab: design summary Commentary It is usually presumed that the detailer would take the design summarised above together with the general arrangement illustrated in Figure 3.3 and detail the slab to normal best practice. The detailer’s responsibilities, standards and timescales should be clearly defined but it would be usual for the detailer to draw and schedule not only the designed reinforcement but all the reinforcement required to provide a compliant and buildable solution. The work would usually include checking the following aspects and providing appropriate detailing : Minimum areas t Curtailment lengths t Anchorages t This publication is available for purchase from www.concretecentre.com S|e|s 45 45 3? Cont|nuous one-wey so||d s|e| Laps t U-bars t Rationalisation t Critical dimensions t Details and sections t The determination of minimum reinforcement areas, curtailment lengths, anchorages and laps using the principles in Eurocode 2 is shown in detail in the following calculations. In practice these would be determined from published tables of data or by using reference texts [8, 9] . Nonetheless the designer should check the drawing for design intent and compliance with standards. It is therefore necessary for the designer to understand and agree the principles of the detailing used. 3.2.10 Detailing checks a) Minimum areas Minimum area of longitudinal tension (flexural) reinforcement A s,min = 0.26(f ctm f /f yk f ) b t bb d ≥ 0.0013 b t bb d where b t bb = width of tension zone f ctm f = 0.30 × f ck f 0.667 A s,min = 0.26 × 0.30 × 30 0.667 × 1000 × 144/500 = 216 mm 7 2 /m (r = 0.15%) r = H12 @ 225 B1 OK Cl. 9.3.1.1, 9.2.1.1 Table 3.1 Secondary (transverse reinforcement) Minimum 20% A s,req 20% A s,req = 0.2 × 502 = 100 mm q 2 /m Consider A s,min to apply as before. A s,min = 216 mm 2 /m Try H10 @ 350 B2 (224 mm 2 /m) Cl. 9.3.1.1(2) SMDSC [9] Check edge. Assuming partial fixity exists at edges, 25% of A s is required to extend 0.2 × the length of the adjacent span. A s,req = 25% × 639 = 160 mm q 2 /m A s,min as before = 216 mm 2 /m = Use H10 @ 350 (224 mm 2 /m) U-bars at edges Cl. 9.3.1.2(2) Cl. 9.3.1.1, 9.2.1.1 This publication is available for purchase from www.concretecentre.com 46 Curtail 0.2 × 5975 = 1195 mm, say 1200 mm measured from face of support ‡ . Cl. 9.3.1.2(2) Maximum spacing of bars Maximum spacing of bars < 3h < 400 mm OK Cl. 9.3.1.1.(3) Crack control As slab < 200 mm, measures to control cracking are unnecessary. Cl. 7.3.3(1) However, as a check on end span: Loading is the main cause of cracking, = use Table 7.2N or Table 7.3N for w max ww = 0.4 mm and s s s = 241 MPa (see deflection check). Max. bar size = 20 mm or max. spacing = 250 mm = H12 @ 150 B1 OK. Cl. 7.3.3(2), 7.3.1.5 Table 7.2N & interpolation, Table 7.3N & interpolation End supports: effects of partial fixity Assuming partial fixity exists at end supports, 15% of A s is required to extend 0.2 × the length of the adjacent span. A s,req = 15% × 639 = 96 mm q 2 /m But, A s,min as before = 216 mm 2 /m (r = 0.15%) r One option would be to use bob bars, but choose to use U-bars Try H12 @ 450 (251 mm 2 /m) U-bars at supports pp Cl. 9.3.1.2(2) Cl. 9.3.1.1, 9.2.1.1 Curtail 0.2 × 5975 = say, 1200 mm measured from face of support. ‡ Cl. 9.3.1.2(2) b) Curtailment i) End span, bottom reinforcement Assuming end support to be simply supported, 50% of A s should extend into the support. 50% × 639 = 320 mm 2 /m Try H12 @ 300 (376 mm 2 /m) at supports pp Cl. 9.3.1.2(1) In theory, 50% curtailment of reinforcement may take place a l from where the moment of resistance of the section with the remaining 50% would be adequate to resist the applied bending moment. In practice, it is usual to determine the curtailment distance as being a l from where M Ed = M Ed,max /2. Cl. 9.3.1.2(1) Note, 9.2.1.3 (2) ‡ Detail MS2 of SMDSC [9] , suggests 50% of T1 legs of U-bars should extend 0.3l (= say 1800 mm) from face of support by placing U-bars alternately reversed. This publication is available for purchase from www.concretecentre.com 47 3? Cont|nuous one-wey so||d s|e| a) Load arrangement rr Tensile for TT ce in reinforcement, F s F 633 50% 50% 633 (say 500) Tensile r TT esistance of reinforcement X M Ed,max M Edx = R A X – X nX 2 XX /2 M Edx /z l bd l bd n A B A B A B A B b) Bending moment M Edx c) Tensile TT force in bottom reinforcement d) Curtailment of bottom reinforcement 987 987 (say 850) Figure 3.5 Curtailment of bottom reinforcement: actions, bending moments, forces in reinforcement and curtailment Thus, for a single simply supported span supporting a UDL of n, M Ed,max = 0.086nl 2 ; R A = 0.4nl At distance, X, from end support, moment, XX M Ed @X = X R A X – X nX 2 /2 = when M@X = X M Ed,max /2: 0.086nl 2 /2 = 0.4nlX – X nX 2 /2 This publication is available for purchase from www.concretecentre.com 48 Assuming X = X xl 0.043nl 2 = 0.4nlxl – l nx 2 xx l 2 /2 0.043 = 0.4x – x x 2 xx /2 0 = 0.043 – 0.4x + x x 2 xx /2 x = 0.128 or 0.672, say 0.13 and 0.66 x = at end support 50% moment occurs at 0.13 x span 0.13 × 5975 = 777 mm Shift rule: for slabs, a l may be taken as d (= 144 mm), d = curtail to 50% of required reinforcement at 777 – 144 = 633 mm from centreline of support. Say 500 mm from face of support A pp Cl. 9.2.1.3(2), 6.2.2(5) = in end span at 1st internal support 50% moment occurs at 0.66 × span 0.66 × 5975 = 3944 mm Shift rule: for slabs a l may be taken as d (= 144 mm), d = curtail to 50% of required reinforcement at 3944 + 144 = 4088 mm from support A or 5975 – 4088 = 987 mm from centreline of support B. Say 850 mm from face of support B pp Cl. 9.2.1.3(2), 6.2.2(5) ii) 1st interior support, top reinforcement Presuming 50% curtailment of reinforcement is required this may take place a l from where the moment of resistance of the section with the remaining 50% would be adequate. However, it is usual to determine the curtailment distance as being a l from where M Ed = M Ed,max /2. Cl. 9.3.1.2(1) Note, 9.2.1.3(2) Thus, for the 1st interior support supporting a UDL of n, M Ed,maxT = 0.086 T nl 2 ; R B = 0.6nl At distance Y from end support, moment, Y M Ed @Y = Y M Ed,maxT – T R A Y + Y nY 2 /2 = when M@Y = Y M Ed,maxT /2 0.086nl 2 /2 = 0.086nl 2 – 0.6nlY + Y nY 2 /2 Assuming Y = Y yl 0.043nl 2 = 0.086nl 2 – 0.6nlyl + l ny 2 l 2 /2 0 = 0.043 − 0.6y + y y 2 /2 y = 0.077 y (or 1.122), say 0.08 = at end support 50% moment occurs at 0.08 × span 0.08 × 5975 = 478 mm Shift rule: for slabs, a l may be taken as d 144 mm d = curtail to 50% of required reinforcement at 478 + 144 = 622 mm from centreline of support. 50% of reinforcement may be curtailed at, say, 600 mm from either face of support B pp Cl. 9.2.1.3(2), 6.2.2(5) This publication is available for purchase from www.concretecentre.com 49 3? Cont|nuous one-wey so||d s|e| 100% curtailment may take place a l from where there is no hogging moment. Thus, when M@Y = M Ed,maxT /2 0 = 0.086nl 2 – 0.6nlY + nY 2 /2 Assuming Y = yl 0 = 0.086 – 0.6y + y 2 /2 y = 0.166 (or 1.034), say 0.17 = at end support 50% moment occurs at 0.17 × span 0.17 × 5975 = 1016 mm Shift rule: for slabs, a l may be taken as d = curtail to 100% of required reinforcement at 1016 + 144 = 1160 mm from centreline of support. 100% of reinforcement may be curtailed at, say, 1100 mm from either face of support B. iii) Support B bottom steel at support At the support 25% of span steel required Cl. 9.3.1.1(4), 9.2.1.5(1), 9.2.1.4(1) 0.25 × 639 = 160 mm 2 A s,min as before = 216 mm 2 /m For convenience use H12 @ 300 B1 (376 mm 2 /m) Cl. 9.3.1.1, 9.2.1.1 c) Anchorage at end support As simply supported, 50% of A s should extend into the support. This 50% of A s should be anchored to resist a force of Cl. 9.2.1.2(1) & Note, 9.2.1.4(2) F E = V Ed × a l /z where Exp. (9.3) V Ed = the absolute value of the shear force a l = d, where the slab is not reinforced for shear z = lever arm of internal forces F E = 29.4 × d/0.95 ‡ d = 30.9 kN/m Cl. 9.2.1.3(2) Anchorage length, l bd : Cl. 8.4.4 l bd = al b,rqd ≥ l b,min Exp. (8.4) where a = conservatively 1.0 l b,rqd = basic anchorage length required = (f/4) (s sd /f bd ) Exp. (8.3) where f = diameter of the bar = 12 mm s sd = design stress in the bar at the ultimate limit state = F E /A s,prov = 30.9 × 1000/376 = 81.5 MPa ‡ Maximum z = 0.947 at mid-span and greater towards support. This publication is available for purchase from www.concretecentre.com 50 f bd = ultimate bond stress = 2.25 n 1 n 2 f ct,d Cl. 8.4.2(2) where n 1 = 1.0 for ‘good’ bond conditions and 0.7 for all other conditions = 1.0 n 2 = 1.0 for bar diameter ≤ 32 mm f ct,d = design tensile strength = a ct f ct,k / g C . For f ck = 30 MPa = 1.0 × 2.0/1.5 = 1.33 MPa Cl. 3.1.6(2) & NA, Tables 3.1 & 2.1N =f |d = 2.25 × 1.33 = 3.0 MPa l b,rqd = (12/4) (81.5/1.33) = 183 mm l b,min = max(10d, 100 mm) = 120 mm l bd = 183 mm measured from face of support By inspection, using U-bars, OK Exp. (8.6) Fig. 9.3 d) Laps Lap H12 @ 300 U-bars with H12 @ 150 straights. Tension lap, l 0 = a 1 a 2 a 3 a 5 a 6 l b,rqd a l 0min Exp. (8.10) where a 1 = 1.0 (straight bars) a 2 = 1 − 0.15 (c d − f)/f Table 8.2 where c d = min(pitch, side cover or cover) = 25 mm Fig. 8.4 f = bar diameter = 12 mm a 2 = 0.84 a 3 = 1.0 (no confinement by reinforcement) a 5 = 1.0 (no confinement by pressure) Table 8.2 a 6 = 1.5 Table 8.3 l b,rqd = (f/4) s sd /f bd Exp. (8.3) where s sd = the design stress at ULS at the position from where the anchorage is measured. Assuming lap starts 500 mm from face of support (587.5 mm from centreline of support): M Ed = 29.5 × 0.59 − 12.3 × 0.59 2 /2 = 15.2 kNm s sd = M Ed /(A s z) = 15.2 × 10 6 /(376 × 144/0.95) = 267 MPa f bd = ultimate bond stress = 2.25 n 1 n 2 f ct,d Cl. 8.4.2(2) This publication is available for purchase from www.concretecentre.com 51 3? Cont|nuous one-wey so||d s|e| where n 1 = 1.0 for ‘good’ conditions n 2 = 1.0 for f < 32 mm f ct,d = a ct f ct,k /g C where a ct = 1.0 f ct,k = 2.0 g C = 1.5 =f |d = 2.25 × 2.0/1.5 = 3.0 MPa Cl. 3.1.6 (2) & NA Table 3.1 Table 2.1N & NA l b,rqd = (f/4) s sd /f bd = (12/4) × (267/3) = 267 mm l 0min b = max[0.3 a 6 l b,rqd ; 15f/ 200 mm] = max[0.3 × 1.5 × 229; 15 × 12; 200] = max[124; 180; 200] = 200 mm = l 0 = a 1 a 2 a 3 a 5 a 6 l b,rqd ≥ l 0min = 1.0 × 0.84 × 1.0 × 1.0 × 1.5 × 329 ≥ 200 = 414 mm Exp. 8.6 But good practice suggests minimum lap of max[tension lap; 500] = lap with bottom reinforcement = 500 mm starting 500 from face of support. 3.2.11 Summary of reinforcement details SMDSC [9] : MS2 A A 500 1200 H12 U-bars @ 300 H12 @ 175 T1 H12 @ 150 H12 @ 300 H12 @ 225 500 600 600 200 200 200 500 500 500 500 350 350 Figure 3.6 Continuous solid slab: reinforcement details 500 1200 H10 @ 350 B2 Figure 3.7 Section A–A showing reinforcement details at edge This publication is available for purchase from www.concretecentre.com 52 3.3 Continuous ribbed slab lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Continuous ribbed slab This continuous 300 mm deep ribbed slab has spans of 7.5 m, 9.0 m and 7.5 m and is required for an office to support a variable action of 5 kN/m 2 . It is supported on wide beams that are the same depth as the slab designed in Section 4.3. One hour fire resistance is required: internal environment. The ribs are 150 mm wide @ 900 mm centres. Links are required in span to facilitate prefabrication of reinforcement. Assume that partitions are liable to be damaged by excessive deflections. In order to reduce deformations yet maintain a shallow profile use f ck = 35 MPa and f yk = 500 MPa. g k = 4.17 kN/m 2 g k = 4.3 kN/m 2 q k = 5.0 kN/m 2 A B 7500 550 1000 1000 1000 9000 7500 C D 550 1000 Figure 3.8 Continuous ribbed slab example p Notes on ribbed slab design There are various established methods for analysing ribbed slabs and dealing with the solid areas: t Using UDLs simplifies the analysis and remains popular. One method is to ignore the weight of the solid part of the slab in the analysis of the ribbed slab. (The weight of the solid area is then added to the loads on the supporting beam). This ignores the minor effect the solid areas have on bending in the ribbed slab. t Alternatively the weight of the solid part of the slab is spread as a UDL over the whole span. This is conservative both in terms of moment and shears at solid/shear interfaces but underestimates hogging in internal spans. The advent of computer analysis has made analysis using patch loads t more viable and the resulting analysis more accurate. t The ribbed part of the slab may be designed to span between solid areas. (The ribs span d/2 into the solid areas, which are assumed to act as beams in the orthogonal direction.) However, having to accommodate torsions induced in supporting beams and columns usually makes it simpler to design from centreline of support to centreline of support. Analysis programs can cope with the change of section and therefore t change of stiffness along the length of the slab. Moments would be attracted to the stiffer, solid parts at supports. However, the difference in stiffness between the ribbed and the solid parts is generally ignored. This publication is available for purchase from www.concretecentre.com 53 In line with good practice analysis, this example is carried out using centreline of support to centreline of support and patch loads ‡ . Constant stiffness along the length of the slab has been assumed. 300 200 200 C L 1000 550 100 A A C L Figure 3.9 Long section through slab 150 150 750 Figure 3.10 Section A–A: section through ribbed slab 3.3.1 Actions Permanent: UDL kN/m 2 Self-weight: kN/m 2 Rib 0.15 × 0.2 × 25/0.9 = 0.833 Slope 2 × (1/2) × 0.2/10 × 0.2 × 25/0.9 = 0.112 Slab 0.1 × 2.5 = 2.500 Cross rib 0.19 × 0.71 × 0.2 × 25/(0.9 × 7.5) = 0.100 Total self-weight = 3.545 ≈ 3.55 Ceiling = 0.15 Services = 0.30 Raised floor = 0.30 Total permanent actions g k = 4.30 ‡ In this case, assuming the patch load analysis is accurate, taking the weight of solid area to be spread over the whole span would overestimate span and support moments by 6–8% and shears at the solid/rib interface by 8–9%. Ignoring the weight of the solid area in the analysis of this ribbed slab would lead to underestimates of span moments by 1%, support moments by 3% and no difference in the estimation of shear at the solid shear interface. The latter may be the preferred option. 33 Cont|nuous r|||ed s|e| This publication is available for purchase from www.concretecentre.com 54 Permanent: patch load Extra over solid in beam area as patch load (0.2 × 25 – 0.833) = 4.167 g k ≈ 4.17 Variable Imposed = 4.00* Allowance for partitions = 1.00* Total variable action g k = 5.00 3.3.2 Cover Nominal cover, c nom : c nom = c min + Dc dev Exp. (4.1) where c min = max(c min,b ; c min,dur ) where c min,b = minimum cover due to bond = diameter of bar. Assume 20 mm main bars and 8 mm links Cl. 4.4.1.2(3) c min,dur = minimum cover due to environmental conditions. Assuming XC1 and C30/37 concrete, c min,dur = 15 mm r Table 4.1. BS 8500-1: Table A4 Dc dev = allowance in design for deviation. Assuming no measurement of cover Dc dev = 10 mm Cl. 4.4.1.2(3) = c nom = 20 + 10 to main bars or = 15 + 10 to links = critical Fire: Check adequacy of section for REI 60. EC2-1-2: 5.7.5(1) Minimum slab thickness, h s = 80 mm OK EC2-1-2: Table 5.8 Axis distance required Minimum rib width b min = 120 mm with a = 25 mm or b min = 200 mm with a = 12 mm EC2-1-2: Table 5.6 = at 150 mm wide (min.) a = 20 mm By inspection, not critical. Use 25 mm nominal cover to links 3.3.3 Load combination and arrangement Ultimate load, n: By inspection, Exp. (6.10b) is critical n slab = 1.25 × 4.30 + 1.5 × 5.0 = 13.38 kN/m 2 n solid areas = 1.25 × (4.30 + 4.17) + 1.5 × 5.0 = 18.59 kN/m 2 Fig. 2.5 Fig. 2.5 EC0: Exp. (6.10b) *Client requirements. See also BS EN 1991–1–1, Tables 6.1, 6.2, Cl. 6.3.2.1(8) & NA. This publication is available for purchase from www.concretecentre.com 55 Arrangement: Choose to use all-and-alternate-spans-loaded. Cl. 5.1.3(1) & NA option b 3.3.4 Analysis Analysis by computer, includes 15% redistribution at support and none in the span. § EC0: A1.2.2 & NA, 5.3.1 (6) 100 90.7 kNm/m 90.7 kNm/m – 61.1 kNm/m – 65.3 kNm/m – 65.3 kNm/m 80 60 40 20 0 –20 –40 –60 –80 A B C D a) Elastic moments 77.1 kNm/m 77.1 kNm/m –61.7 kNm/m –61.7 kNm/m –55.9 kNm/m 100 80 60 40 20 0 –20 –40 –60 –80 A B C D b) Redistributed envelope Figure 3.11 Bending moment diagrams 33 Cont|nuous r|||ed s|e| § Note 1: A ribbed slab need not be treated as discrete elements provided rib spacing ≤ 1500 mm, depth of the rib ≤ 4 × its width, the flange is > 0.1 × distance between ribs and transverse ribs are provided at a clear spacing not exceeding 10 × overall depth of the slab. Note 2: As 7.5 m < 85% of 9.0 m, coefficients presented in Concise Eurocode 2 [5] are not applicable. This publication is available for purchase from www.concretecentre.com 56 A B C D 0 80 40 42.5 kN/m – 42.5 kN/m 63.2 kN/m – 63.2 kN/m 63.5 kN/m – 63.5 kN/m – 80 – 40 At solid/rib interface: AB @ 550 mm from A M Ed (sagging) V Ed = 20.4 kNm/m z 18.3 kNm/rib = 32.5 kN/m z 29.3 kN/rib BA @1000 mm from B M Ed (hogging) V Ed = 47.1 kNm/m z 42.4 kNm/rib = 45.4 kN/m z 40.9 kN/rib BC @ 1000 mm from B M Ed (hogging) V Ed = 43.0 kNm/m z 38.7 kNm/rib = 45.1 kN/m z 40.6 kN/rib Symmetrical about centreline of BC. 3.3.5 Flexural design, span A–B a) Span A–B: Flexure M Ed = 61.7 kNm/m = 55.5 kNm/rib K = M Ed /bd 2 f ck where b = 900 mm d = 300 − 25 – 8 – 20/2 = 257 assuming 8 mm link at H20 in span f ck = 35 MPa =K = 55.5 × 10 6 /(900 × 257 2 × 35) = 0.027 K' = 0.207 or restricting x/d to 0.45 K' = 0.168 K ≤ K' = section under-reinforced and no compression reinforcement required. Appendix A1 This publication is available for purchase from www.concretecentre.com 57 33 Cont|nuous r|||ed s|e| z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d = (257/2) (1 + 0.951) ≤ 0.95 × 257 = 251 ≤ 244 = z = 244 mm Appendix A1 But z = d – 0.4x = x = 2.5(d − z) = 2.5(257 − 244) = 33 mm = By inspection, neutral axis is in flange A s = M Ed /f yd z where | yd = 500/1.15 = 434.8 MPa = 55.5 × 10 6 /(434.8 × 244) = 523 mm 2 /rib Try 2 no. H20/rib (628 mm 2 /rib) Appendix A1 b) Span A–B: Deflection Allowable l/d = N × K × F1 × F2 × F3 Appendix C7 where N = Basic l/d: check whether r > r 0 and whether to use Cl. 7.4.2(2) Exp. (7.16a) or Exp. (7.16b) r 0 = f ck 0.5 /1000 = 35 0.5 /1000 = 0.59% r = A s /A c ‡ = A s,req /[b w d + (b eff – b w )h f ] PD 6687 [6] where b w = min. width between tension and compression chords. At bottom assuming 1/10 slope to rib: = 150 + 2 × (25 + 8 + 20/2)/10 = 159 mm r = 523/(159 ( 257 + (900 − 159) × 100) = 523/114963 = 0.45% r < r 0 = use Exp. (7.16a) N = 11 + 1.5f ck 0.5 r /r 0 + 3.2f ck 0.5 (r /r 0 – 1) 1.5 ] = 11 + 1.5 × 35 0.5 × 0.055/0.045 + 3.2 × 35 0.5 (0.055/0.045 – 1) 1.5 = [11 + 10.8 + 2.0] = 22.8 Exp. (7.16a) K = (end span) 1.3 Table 7.4N & NA, Table NA.5: Note 5 F1 = (b eff /b w = 5.66) 0.8 F2 = 7.0/l eff = 7.0/7.5 = (span > 7.0 m) 0.93 Cl. 7.4.2(2) F3 = 310/ s s ≤ 1.5 Cl. 7.4.2, Exp. (7.17) & NA; Table NA.5 ‡ Section 2.18 of PD 6687 [6] suggests that r in T-beams should be based on the area of concrete above the centroid of the tension steel. This publication is available for purchase from www.concretecentre.com 58 where ‡ s s = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ULS loads) (1/d) = 434.8(523/628) [ (4.30 + 0.3 × 5.0)/13.38] (65.3/61.7 § ) = 434.8 × 0.83 × 0.43 × 1.06 = 164 MPa F3 = 310/s s = 310/164 = 1.89 # but ≤ 1.5, therefore say 1.50 = Permissible l/d = 22.8 × 1.3 × 0.8 × 0.93 × 1.50 = 33.0 Actual l/d = 7500/257 = 29.2 = OK Use 2 no. H20/rib (628 mm 2 /rib) c) Support A (and D): flexure (sagging) at solid/rib interface Reinforcement at solid/rib interface needs to be designed for both moment and for additional tensile force due to shear (shift rule) Cl. 9.2.1.3.(2) M Ed,max = 18.3 kNm/rib V Ed,max = 29.3 kNm/rib At solid/rib interface A s = M Ed /f yd z + DF td /f yd Cl. 9.2.1.3.(2), Fig. 9.2 where z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d where K = M Ed /bd 2 f ck where b = 900 mm d = 300 − 25 – 8 – 25 − 20/2 = 232 assuming 8 mm links and H25B in edge beam f ck = 30 = 18.3 × 10 6 /(900 × 232 2 × 35) = 0.011 ‡ See Appendix B1.5 § In analysis, 15% redistribution of support moments led to redistribution of span moments: d = 61.7/65.3 = 0.94. # Both A s,prov /A s,req and any adjustment to N obtained from Exp. (7.16a) or Exp. (7.16b) is restricted to 1.5 by Note 5 to Table NA.5 in the UK NA. Therefore, 310/ s s is restricted to 1.5. This publication is available for purchase from www.concretecentre.com 59 33 Cont|nuous r|||ed s|e| 25 cover 8 link 8 link 20 bar 25 cover 12 fabric 16 bar 25 cover 8 link 25 bar 16 bar = z = (232/2) (1 + 0.980) ≤ 0.95 × 232 = 230 ≤ 220 = z = 220 mm f yd = 434.8 MPa DF td = 0.5V Ed (cot y – cot a) Appendix A1 Cl. 6.2.3(7), Exp. (6.18) where y = angle between the concrete compression strut and the beam axis. Assume cot y = 2.5 (as a maximum) a = angle between shear reinforcement and the beam axis. For vertical links, cot a = 0 DF td = 1.25V Ed = 1.25 × 29.3 = 36.6 kN A s = 18.3 × 10 6 /(434.8 × 220) + 36.6 × 10 3 /434.8 = 191 + 84 mm 2 = 275 mm 2 =Try 1 no. H20 B in end supports* Cl. 6.2.3(1) Appendix A2 Appendix C, Table C6 Cl. 6.2.3(1) d) Support B (and C) (at centreline of support) M Ed = 77.1 kNm/m = 69.4 kNm/rib K = M Ed /bd 2 f ck where d = 300 − 25 cover − 12 fabric − 8 link − 20/2 = 245 K = 69.4 × 10 6 /(900 × 245 2 × 35) = 0.037 By inspection, K ≤ K' z = (245/2) [1 + (1 − 3.53 K) 0.5 ] ≤ 0·95d = (245/2) (1 + 0.932) < 0.95d = 237 mm A s = M Ed /f yd z = 69.4 × 10 6 /434.8 × 237 = 673 mm 2 /rib * An alternative method would have been to calculate the reinforcement required to resist M Ed at the shift distance, a l , from the interface. This publication is available for purchase from www.concretecentre.com 60 e) Support B (and C): flexure (hogging) at solid/rib interface Reinforcement at solid/rib interface needs to be designed for both moment and for additional tensile force due to shear (shift rule). M Ed,max = 42.4 kNm/rib max. V Ed,max = 40.9 kNm/rib max. Cl. 9.2.1.3.(2) A s = M Ed /f yd z + DF td /f yd where z = (245/2) [1 + (1 − 3.53 K) 0.5 ] ≤ 0·95d where K = M Ed /bd 2 f ck = 42.4 × 106/(150 × 245 2 × 35) = 0.135 Cl. 9.2.1.3.(2) Check K ≤ K' K' = 0.168 for d = 0.85 (i.e. 15% redistribution) = Section under-reinforced: no compression reinforcement required Appendix C, Table C4 Appendix A = z = (245/2) (1 + 0.723) ≤ 232 = 211 mm f yd = 434.8 MPa DF td = 0.5V Ed (cot y – cot a) Cl. 6.2.3(7), Exp. (6.18) where y = angle between the concrete compression strut and the beam axis. Assume cot y = 2.5 (as a maximum) a = angle between shear reinforcement and the beam axis. For vertical links, cot a = 0 DF td = 1.25V Ed = 1.25 × 40.9 = 51.1 kN Cl. 6.2.3(1) Appendix A2; Table C6 Cl. 6.2.3(1) A s = 42.4 × 10 6 /(434.8 × 211) + 51.1 × 10 3 /434.8 = 462 + 117 mm 2 = 579 mm 2 /rib To be spread over b eff where by inspection, b eff = 900. Cl. 9.2.1.2(2) =Centre of support more critical (679 mm 2 /rib required). Cl. 5.3.2.1(3) Top steel may be spread across b eff where b eff = b w + b eff1 + b eff2 ≤ b = b w + 2 × 0.1 × 0.15 × (l 1 + l 2 ) = 150 + 0.03 × (7500 + 9000) ≤ 900 = 645 mm = Use 2 no. H16 above rib and 3 no. H12 between (741 mm 2 /rib) where 2 no. H16 and 2 no. H12 are within b eff Cl. 9.2.1.2(2), 5.3.2 3.3.6 Flexural design, span BC a) Span B–C: Flexure M Ed = 55.9 kNm/m = 50.3 kNm/rib This publication is available for purchase from www.concretecentre.com 61 33 Cont|nuous r|||ed s|e| K = M Ed /bd 2 f ck = 50.3 × 10 6 /900 × 257 2 × 35 = 0.02 i.e. ≤ K’ (as before K’ = 0.168) By inspection, z = 0.95d = 0.95 × 257 = 244 mm By inspection, neutral axis is in flange. A s = M Ed /f yd z = 50.3 × 10 6 /434.8 × 244 = 474 mm 2 Try 2 no. H20/rib (628 mm 2 /rib) b) Span B–C: Deflection Allowable l/d = N × K × F1 × F2 × F3 Section C7 where N = Basic l/d r = 474/(159 (× 257 + (900 − 159) × 100) = 474/114963 = 0.41% r 0 = 0.59% (for f ck = 30) = r < r 0 use Exp. (7.16a) Cl. 7.4.2(2) N = 11 + 1.5 f ck 0.5 r 0 /r + 3.2f ck 0.5 (r 0 /r – 1) 1.5 = 11 + 1.5 × 35 0.5 × 0.055/0.041 + 3.2 × 35 0.5 (0.055/0.041 − 1) 1.5 = 11 + 11.9 + 3.8 = 26.7 Exp. (7.16a) K = (internal span) 1.5 Table 7.4N, & NA, Table NA.5: Note 5 F1 = (b eff /b w = 6.0) 0.8 F2 = 7.0/l eff = 7.0/9.0 = (span > 7.0 m) 0.77 Cl. 7.4.2(2) F3 = 310/s S ≤ 1.5 where s s = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ULS loads) (1/d) = 434.8 × (474/628) [(4.30 + 0.3 × 5.0)/13.38](61.1/55.9) = 434.8 × 0.75 × 0.43 × 1.09 = 153 MPa Cl. 7.4.2, Exp. (7.17) & NA: Table NA.5 F3 = 310/ s s = 310/153 = 2.03 therefore, say = 1.50 ‡ = Permissible l/d = 26.8 × 1.5 × 0.8 × 0.77 × 1.50 = 37.1 Actual l/d = 9000/257 = 35 = OK =Use 2 H20/rib (628 mm 2 /rib) NA, Table NA.5: Note 5 ‡ Both A s,prov /A s,req and any adjustment to N obtained from Exp. (7.16a) or Exp. (7.16b) is restricted to 1.5 by Note 5 to Table NA.5 in the UK NA. This publication is available for purchase from www.concretecentre.com 62 3.3.7 Design for shear 10 C L C L b = 150 1 a) Support A (and D) at solid/rib interface Shear at solid/rib interface = 29.3 kN/rib Taking solid area as the support, at d from face of support V Ed = 29.3 − 0.232 × 0.90 × 13.38 = 26.5 kN/rib Cl. 6.2.1(8) Cl. 6.2.2(1) & NA Resistance V Rd,c = (0.18/g C )k (100 r l f ck ) 0.333 b w d where g C = 1.5 k = 1 + (200/d) 0.5 ≤ 2 = 1 + (200/257) 0.5 = 1.88 r l = A sl /b w d where A sl = assume only 1 H20 anchored = 314 mm 2 b w = min. width between tension and compression chords. At bottom assuming 1/10 slope to rib: = 150 + 2 × (25 + 8 + 20/2)/10 = 159 mm d = 257 mm as before r l = 314/(159 × 257) = 0.0077 f ck = 35 =V Rd,c = (0.18/1.5) 1.88 (100 × 0.0077 × 35) 0.333 × 159 × 257 = 0.68 × 159 × 257 = 27.8 kN/rib = No shear links required. But use nominal links to allow prefabrication. Cl. 6.2.1(5) b) Support B (and C) at solid/rib interface Shear at solid/rib interface = 40.9 kN/rib [max(B A ; B C )] At d from face of support V Ed = 40.9 − 0.245 × 13.38 × 0.9 = 37.9 kN/rib Cl. 6.2.1(8) This publication is available for purchase from www.concretecentre.com 63 33 Cont|nuous r|||ed s|e| Resistance: V Rd,c = (0.18/ g C )k (100 r l f ck ) 0.333 b w d where g C = 1.5 k = 1 + (200/d) 0.5 ≤ 2 = 1 + (200/245)0.5 = 1.90 r l = A sl /b w d where A sl = 2 H16 = 402 mm 2 b w = 159 mm as before d = 245 mm as before r l = 0.0103 f ck = 35 MPa = V Rd,c = (0.18/1.5) 1.9 (100 × 0.0103 × 35) 0.333 × 159 × 245 = 0.75 × 159 × 245 = 29.2 kN/rib = Shear links required. Cl. 6.2.2(1) & NA Shear links required for a distance: (37.9 − 29.2)/(13.38 × 0.9) + 245 = 722 + 245 = 967 mm from interface. Check shear capacity: V Rd,max = a cw b w zvf cd /(cot y + tan y) Exp. (6.9) & NA where a cw = 1.0 b w = 159 mm as before z = 0.9d v = 0.6 (1 − f ck /250) = 0.528 f cd = 35/1.5 = 23.3 MPa y = angle of inclination of strut. Rearranging formula above: (cot y + tan y) = a cw b w zvf cd /V Ed = (1.0 × 159 × 0.9 × 245 × 0.528 × 23.3) 41.6 × 103 = 10.4 By inspection, cot −1 y << 21.8. But cot y restricted to 2.5 and = tan y = 0.4. V Rd,max = 1.0 × 159 × 0.9 × 245 × 0.528 × 20/(2.5 + 0.4) = 127.6 kN = OK Cl. 6.2.3(2) & NA This publication is available for purchase from www.concretecentre.com 64 Shear links: shear resistance with links V Rd,s = (A sw /s) z f ywd cot y ≤ V Rd,max where A sw /s = area of legs of links/link spacing z = 0.9d as before f ywd = 500/1.15 = 434.8 cot y = 2.5 as before = for V Ed ≤ V Rd,s A sw /s ≥ V Ed /z f ywd cot y ≥ 37.9 × 10 3 /(0.9 × 245 × 434.8 × 2.5) ≥ 0.158 Exp. (6.8) Maximum spacing of links = 0.75d = 183 mm = Use H8 @ 175 cc in 2 legs (A sw /s = 0.57) for min. 967 mm into rib Cl. 9.2.2(6) 3.3.8 Indirect supports As the ribs of the slab are not supported at the top of the supporting beam sections (A, B, C, D), additional vertical reinforcement should be provided in these supporting beams and designed to resist the reactions. This additional reinforcement should consist of links within the supporting beams (see Beams design, Section 4.3.9). Cl. 9.2.5, Fig. 9.7 Support A (and D) at solid/rib interface: V Ed = 26.5 kN/rib A s,req = 26.3 × 1000/(500/1.15) = 60 mm 2 This area is required in links within h/6 = 300/6 = 50 mm of the ribbed/solid interface and within h/2 = 300/2 = 150 mm of the centreline of the rib. Fig. 9.7 Support B (and C) at solid/rib interface: V Ed = 37.9 kN/rib A s,req = 37.9 × 1000/(500/1.15) = 87 mm 2 placed similarly 3.3.9 Other checks Check shear between web and flange By inspection, V Ed ≤ 0.4 f ct,d = OK Cl. 6.4.2 (6) & NA This publication is available for purchase from www.concretecentre.com 65 33 Cont|nuous r|||ed s|e| 3.3.10 Summary of design 2H16 + 3H12/rib H8 links in 2 legs @ 175cc A 550 550 1050 7500 f ck = 35 MPa c nom = 25 mm 7500 9000 1000 1000 1050 1050 1000 1000 1050 B C D 2H20/rib 2H20/rib 2H20/rib 2H16 + 3H12/rib Figure 3.15 Summary of design Commentary It is usually presumed that the detailer would take the above design and detail the slab to normal best practice. As stated in Section 3.2.9, the detailer’s responsibilities, standards and timescales should be clearly defined but it would be usual for the detailer to draw and schedule not only the designed reinforcement but all the reinforcement required to provide a buildable solution. The work would usually include checking the following aspects and providing appropriate detailing: Minimum areas t Curtailment lengths t Anchorages t Laps t U-bars t Rationalisation t Details and sections t The determination of minimum reinforcement areas, curtailment lengths and laps using the principles in Eurocode 2 is shown in detail in the following calculations. In practice these would be determined from published tables of data or by using reference texts [12, 21] . Nonetheless the designer should check the drawing for design intent and compliance with standards. It is therefore necessary for the designer to understand and agree the principles of the details used. 3.3.11 Detailing checks a) Minimum areas i) Minimum area of reinforcement in flange A s,min = 0.26 (f ctm /f yk ) b t d ≥ 0.0013 b t d Cl. 9.3.1.1 This publication is available for purchase from www.concretecentre.com 66 where b t = width of tension zone Cl. 9.2.1.1, Exp. (9.1N) f ctm = 0.30 × f ck 0.666 Table 3.1 A s,min = 0.26 × 0.30 × 35 0.666 × 1000 × 100/500 = 166 mm 2 /m (r = 0.17%) = Use A142 in flange (say OK) BS 8666 [19] ii) Secondary reinforcement Not applicable. iii) Maximum spacing of bars Maximum spacing of bars < 3 h < 400 mm By inspection. OK Cl. 9.3.1.1.(3) iv) Crack control Loading is the main cause of cracking = use Table 7.2N or Table 7.3N for Cl. 7.3.3(2) w max = 0.3 mm and max. s s = 200 MPa (see deflection check) Cl. 7.3.1.5 Max. bar size = 25 mm Table 7.2N or max. spacing = 250 mm Table 7.3N OK by inspection v) Effects of partial fixity Assuming partial fixity exists at end supports, 15% of A s is required to extend 0.2 × the length of the adjacent span. A s,req = 15% × 525 = 79 mm 2 /rib For the rib in tension: A s,min = 0.26 × 0.30 × 30 0.666 × 159 × 257/500 = 55 mm 2 /rib Cl. 9.3.1.2(2) b) Curtailment Wherever possible simplified methods of curtailing reinforcement would be used. The following is intended to show how a rigorous assessment of curtailment of reinforcement might be undertaken. i) End support A: bottom steel at support Check anchorage. As simply supported, 25% of A s should be anchored in support. Cl. 9.3.1.1(4), 9.3.1.2(1) & Note, 25% × 595 = 148 mm 2 Use 1 no. H20/rib (314 mm 2 /rib) Cl. 9.2.1.4(1) & NA ii) Check anchorage length Envelope of tensile force: To resist envelope of tensile force, provide reinforcement to a l or l bd Cl. 9.3.1.1(4), 9.2.1.3(1), beyond centreline of support. For members without shear reinforcement, a l = d = 232 By inspection, s sd = 0, l bd = l bd,min = max(10f, 100 mm) Cl. 9.2.1.3(2), 9.2.1.3(3), Fig. 9.2 Cl. 9.2.1.3 iii) Indirect support As anchorage may be measured from face of indirect support, check force to be resisted at solid/rib interface: F s = M Ed /z + F E Cl. 9.3.1.1(4), 9.2.1.4(2), 9.2.1.4(3), Fig. 9.3b This publication is available for purchase from www.concretecentre.com 67 33 Cont|nuous r|||ed s|e| where M Ed = 18.3 kNm/rib z = 220 as before F E = V Ed × a l /z Exp. (9.3) where V Ed = 29.3 kN/rib a l = z cot y/2 = F E = V Ed cot y/2 = 29.3 × 1.25 = 36.6 kN/rib F s = 18.6 × 10 6 /(220 × 10 3 ) + 36.6 = 121.1 kN Cl. 9.2.1.3, Exp. (9.2) iv) Anchorage length: l bd = al b,rqd ≥ l b,min where Cl. 8.4.4, Exp. (8.4) a = conservatively 1.0 l b,rqd = (f/4) (s sd /f bd ) where f = 20 s sd = design stress in the bar at the ULS = 121.1 × 1000/314 = 385 MPa f bd = ultimate bond stress Exp. (8.3) = 2.25 n 1 n 1 f ct,d where n 1 = 1.0 for good bond conditions n 2 = 1.0 for bar diameter ≤ 32 mm Cl. 8.4.2(2) f ct,d = a ct f ct,k /g C = 1.0 × 2.2/1.5 = 1.47 MPa f bd = 2.25 × 1.47 = 3.31 MPa Cl. 3.1.6(2), Tables 3.1, 2.1 & NA = l b,rqd = (20/4) (385/3.31) = 581 mm l b,min = max[10f; 100 mm] = 200 mm = l bd = 581 mm measured from solid/rib intersection. i.e. 31 mm beyond centreline of support ‡ . Fig. 9.3 v) End support A: top steel Assuming partial fixity exists at end supports, 15% of A s is required to extend at least 0.2 × the length of the adjacent span § . Cl. 9.3.1.2(2) A s,req = 15% × 525 = 79 mm 2 /rib A s,min = 0.26 × 0.30 × 35 0.666 × 159 × 257/500 = 68 mm 2 /rib Cl. 9.3.1.1 Cl. 9.2.1.1(1), Use 2 no. H12 T1/rib in rib and 2 no. H10 T1/rib between ribs (383 mm 2 /rib) Exp. (9.1N) ‡ Whilst this would comply with the requirements of Eurocode 2, it is common practice to take bottom bars 0.5 × a tension lap beyond the centreline of support (= 250 mm beyond the centreline of support; see model detail MS1 in SMDSC [9] ). § It is usual to curtail 50% of the required reinforcement at 0.2l and to curtail the remaining 50% at 0.3l or line of zero moment (see model detail MS2 in SMDSC [9] ). This publication is available for purchase from www.concretecentre.com 68 vi) Support B (and C): top steel At the centreline of support (2 no. H16 T + 3 no. H12 T)/rib are required. The intention is to curtail in two stages, firstly to 2 no. H16 T/rib then to 2 no. H12 T/rib. Curtailment of 2 no. H16 T/rib at support (capacity of 2 no. H12 T/rib + shift rule): Assume use of 2 no. H12 T throughout in midspan: Assuming z = 211 mm as before, M R2H12T = 2 × 113 × 434.8 × 211 = 20.7 kNm/rib (23.0 kNm/m) (Note: section remains under-reinforced) From analysis M Ed = 23.0 kNm/m occurs at 2250 mm (towards A) and 2575 mm (towards B). Shift rule: a l = z cot y/2 Assuming z = 211 mm as before a l = 1.25 × 211 = 264 mm = 2 no. H12 T are adequate from 2250 + 264 = 2513 mm from B towards A and 2575 + 263 = 2838 mm from B towards C. = Curtail 2 no. H16 T @ say 2600 from B A and 2850 from B C Curtailment of 3 no. H12 T/rib at support (capacity of 2 no. H16 T/rib + shift rule): M R2H16T = 2 × 201 × 434.8 × 211 = 36.9 kNm/rib (41.0 kNm/m) (Note: section remains under-reinforced) From analysis M Ed = 41.0 kNm/m occurs at 1310 mm (towards A) and 1180 mm (towards C). Shift rule: a l = 263 mm as before = 2 no. H16 T are adequate from 1310 + 263 = 1573 mm from B towards A and 1180 + 263 = 1443 mm from B towards C. = Curtail 3 no. H12 at say 1600 from B (or C). (See Figure 3.16) vii) Support B (and C): bottom steel at support At the support 25% of span steel required 0.25 × 628 = 157 mm 2 Cl. 9.3.1.1(4), 9.2.1.5(1), 9.2.1.4(1) Try 1 no. H16 B/rib (201) This reinforcement may be anchored into indirect support or carried through. Fig. 9.4 This publication is available for purchase from www.concretecentre.com 69 33 Cont|nuous r|||ed s|e| 2H12/rib 2H16/rib 2H16 + 3H12 per rib ‘Shift’ moment a) Design moments and moment resistance b) Curtailment of reinforcement 264 264 264 2H12 3H12 2H12 2H16 1600 1600 2600 2850 264 2250 2575 1310 1180 B A C l bd a 1 = 264 M Ed = 77.1 x 0.9 = 60.4 kNm/rib M Ed = 20.7 kNm/rib M Ed = 36.79 kNm/rib M Ed M R = A s (f yk /g S )z p p viii) Support B (and C): bottom steel curtailment BA and BC To suit prefabrication 2 no. H20/rib will be curtailed at solid/rib interface, 1000 mm from B A (B towards A) and B C . From analysis, at solid/rib interface sagging moment = 0. From analysis, at a 1 from solid/rib interface, i.e. at 1000 + 1.25 × 244 = 1303 mm at 1305 mm from B A sagging moment = say 5 kNm/rib at 1305 mm from B C sagging moment = 0 Use 1 no. H16 B/rib (201) c) Laps At A B , check lap 1 no. H20 B to 2 no. H20 B in rib full tension lap: l 0 = a 1 a 6 l b,rqd > l 0,min Exp. (8.10) where a 1 = 1.0 (c d = 45 mm, i.e. < 3f) a 6 = 1.5 (as > 50% being lapped) l b,rqd = (f/4) (s sd /f bd ) where f = 20 s sd = 434.8 f bd = 3.0 MPa as before Table 8.2 This publication is available for purchase from www.concretecentre.com 70 l 0,min = max. 10f or 100 = 200 l 0 = 1.0 × 1.5 × (20/4) × 434.8/3.0 Exp. (8.6) = 1087 mm, say = 1200 mm SMDSC [9] At B A and B C , check lap 2 no. H12 T to 2 no. H16 T in rib – full tension lap: l 0 = a 1 a 6 l b,rqd > l 0,min where a 1 = 0.7 (c d = 45 mm, i.e. > 3f) a 6 = 1.5 (as > 50% being lapped) l b,rqd = (f/4) (s sd /f bd ) where f = 20 s sd = 434.8 f bd = 2.1 (3.0 MPa as before but n 1 = 0.7 for “not good bond conditions”) l 0,min = max. 10f or 100 = 120 Exp. (8.10) Table 8.2 Cl. 8.4.2 l 0 = 0.7 × 1.5 × (12/4) × 434.8/2.1 = 651 mm, say = 700 mm But to aid prefabrication take to solid/rib intersection 1000 mm from centre of support. SMDSC [9] At B A and B C , check lap 1 no. H16 B to 2 no. H20 B in rib: By inspection, nominal say, 500 mm SMDSC [9] d) RC detail of ribbed slab Links not shown for clarity. Cover 25 mm to links. 200 200 1500 2H12T in rib and 2H1OT between 2H16 + 3H12/rib 2H16 + 3H12/rib A143 fabric 2H12 3H12 2H16 2H12 3H12 2H16 2H12 150 550 500 7500 9000 500 1000 1000 500 1000 1000 1200 2H12 + 2H1O 2H16 3H12 2H12 2H12T/rib 2H12T 1H20B 2H20B/rib 2H20B/rib 1H16B/rib 1H16B/rib 1000 1000 1000 600 600 600 1000 1250 1250 Figure 3.17 Curtailment of flexural reinforcement in ribbed slab This publication is available for purchase from www.concretecentre.com 71 33 Cont|nuous r|||ed s|e| The slab is for an office where the specified load is 1.0 kN/m 2 for finishes and 4.0 kN/m 2 imposed (no partitions). Perimeter load is assumed to be 10 kN/m. Concrete is C30/37. The slab is 300 mm thick and columns are 400 mm square. The floor slabs are at 4.50 m vertical centres. A 2 hour fire rating is required. E D C A 4.0 8.0 9.6 200 x 200 hole 200 x 200 hole 300 mm flat slabs All columns 400 mm sq. 8.6 8.0 4.0 4.0 6 .0 1 2 3 B Bb Figure 3.18 Part plan of flat slab p 3.4.1 Actions kN/m 2 Permanent: EC1-1-1: Table A1 Self-weight 0.30 × 25 = 7.5 Finishes = 1.0 Total g k = 8.5 Variable: Offices ¡ k = 4.0 ‡ ‡ Client requirement. See also BS EN 1991–1–1, Tables 6.1, 6.2, Cl. 6.3.2.1(8) & NA. 3.4 Flat slab ¹h|s exemp|e |s ¦or the des|gn o¦ e re|n¦orced concrete ¦|et s|e| w|thout co|umn heeds ¹he s|e| |s pert o¦ e |erger ¦|oor p|ete end |s te|en ¦rom Oa/Je |o |/e Je·/qn onJ con·|·ac|/on o| ·e/n|o·ceJ conc·e|e |/o| ·/o/· |?/| , where ¦|n|te e|ement ene|ys|s end des|gn to lurocode ? |s |||ustreted As w|th the Oa/Je, gr|d ||ne C w||| |e des|gned |ut, ¦or the se|e o¦ |||ustret|on, coe¦¦|c|ents w||| |e used to este|||sh des|gn moments end sheers |n th|s cr|t|ce| eree o¦ the s|e| lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Flat slab This publication is available for purchase from www.concretecentre.com 72 3.4.2 Cover c nom : c nom = c min + Dc dev where c min = max[c min,b ; c min,dur ; 10 mm] where c min,b = 20 mm, assuming 20 mm diameter reinforcement c min,dur = 15 mm for XC1 and using C30/37 Dc dev = 10 mm Exp. (4.1) Cl. 4.4.1.2(3) Table 4.1. BS 8500-1: Table A4. Fire: For 2 hours resistance, a min = 35 mm = not critical = c nom = 20 + 10 = 30 mm EC2-1-2: Table 5.9 3.4.3 Load combination and arrangement q k = 4.0 kN/m 2 g k = 8.5 kN/m 2 9600 8600 2 3 1 Figure 3.19 Panel centred on grid C Ultimate load, n: By inspection, Exp. (6.10b) is critical. n = 1.25 × 8.50 + 1.5 × 4.0 = 16.6 kN/m 2 Fig. 2.5 EC0: Exp. (6.10b) Arrangement: Choose to use all-and-alternate-spans-loaded load cases and coefficients ‡ . Cl. 5.1.3(1) & NA: Table NA.1 (option b) 3.4.4 Analysis grid line C Consider grid line C as a bay 6.0 m wide. (This may be conservative for grid line C but is correct for grid line D etc.) M Ed Effective spans: 9600 – 2 × 400/2 + 2 × 300/2 = 9500 mm 8600 – 2 × 400/2 + 2 × 300/2 = 8500 mm Cl. 5.3.2.2(1) Check applicability of moment coefficients: 8500/9500 = 0.89 = as spans differ by less than 15% of larger span, coefficients are applicable. Tables C2 & C3 ‡ The all-spans-loaded case with 20% redistribution of support moments would also have been acceptable but would have involved some analysis. The use of Table 5.9 in BS EN 1992–1–2 (Fire resistance of solid flat slabs) is restricted to where redistribution does not exceed 15%; the coefficients presume 15% redistribution at supports. Cl. 5.3.1 & NA Table C3 This publication is available for purchase from www.concretecentre.com 73 34 l|et s|e| As two span, use table applicable to beams and slabs noting increased coefficients for central support moment and shear. Table C3 Design moments in bay. Spans: M Ed = (1.25 × 8.5 × 0.090 + 1.5 × 4.0 × 0.100)× 6.0 × 9.5 2 = 842.7 kNm Support: M Ed = 16.6 × 0.106 × 6.0 × 9.5 2 = 952.8 kNm C D Column strip Column strip Column strip Column strip Middle strip Middle strip 6000 1500 1500 1500 15001500 1500 3000 1500 1 2 Figure 3.20 Column and middle strips p Apportionment of moments between column strips and middle strips: Apportionment (as %) Column strip Middle strip –ve (hogging) Long span = 70% § Short span = 75% Long span = 30% Short span = 25% +ve (sagging) 50% 50% Table I.1; CS Flat slab guide [27] Table I.1 NA.3 [1a] ; Fig. I.1 Parallel to grid C, column strip is l y /2 = 3 m wide. The middle strip is also 3 m wide. Long span moments: M Ed Column strip, 3 m wide Middle strip, 3 m wide –ve (hogging) 0.70 × 952.8/3.0 = 222.3 kNm/m 0.30 × 952.8/3.0 = 95.3 kNm/m +ve (sagging) 0.50 × 842.7/3.0 = 140.5 kNm/m 0. 50 × 842.7/3.0 = 140.5 kNm/m § The Concrete Society’s TR 64 [27] recommends a percentage, k 1 , based on l y /l z Assuming l y /l z = 1.5 the distribution of moments in the long span between column strips and middle strips is given as 70% and 30%. This publication is available for purchase from www.concretecentre.com 74 Punching shear force, V Ed VV : At C2, V Ed VV = 16.6 × 6.0 × 9.6 ‡ × 0.63 × 2 = 1204.8 kN Table C3 Table C3 At C1 (and C3) V Ed VV = 16.6 × 6.0 × 9.6 × 0.45 + (10 + 0.2 × 0.3 × 25) § × 1.25 × 6.0 = 516.5 kN Table C3 Table C3 3.4.5 Design grid line C Effective depth, d: d = 300 − 30 − 20/2 = 260 mm d a) Flexure: column strip and middle strip, sagging M Ed = 140.5 kNm/m K = M Ed /bd 2 f ck f = 140.5 × 10 6 /(1000 × 260 2 × 30) = 0.069 z/ zz d = 0.94 d Table C5 Table C5 z = 0.94 × 260 = 244 mm A s = M Ed /f yd ff z = 140.5 × 10 z 6 /(244 × 500/1.15) = 1324 mm 2 /m (r = 0.51%) r Try H20 @ 200 B1 (1570 mm 2 /m) b) Deflection: column strip and middle strip Check span-to-effective-depth ratio. Appendix B Appendix B Allowable l/ ll d = d N × N K × F1 × F2 × F3 where Cl. 7.4.2(2) Appendix C Appendix C N = 20.3 ( N r = 0.51%, r f ck f = 30) Tables C10–C13 Tables C10–C13 K = 1.2 (flat slab) F1 = 1.0 (b eff /b w bb = 1.0) w F2 = 1.0 (no brittle partitions) # F3 = 310/s s s ≤ 1.5 Cl. 7.4.2, Exp. (7.17) Table 7.4N, & NA, Table NA.5 Note 5 Fig. C3 Fig. C3 where* s s s = s su s (A s,req / q A s,prov ) 1/d where s su s = (500/1.15) × (8.5 + 0.3 × 4.0)/16.6 = 254 MPa (or ≈ 253 MPa; from Figure C3 G k /Q k = 2.1, c 2 cc = 0.3 and g G gg = 1.25) d = redistribution ratio = 1.03 d = s s s ≈ 253 × (1324/1570)/1.03 = 207 = F3 = 310/207 = 1.50 † = Allowable l/ ll d = 20.3 × 1.2 × 1.50 = 36.5 d Fig. C14 Fig. C14 ‡ As punching shear force (rather than a beam shear force) ‘effective’ span is not appropriate. § Cladding and strip of slab beyond centre of support. # Otherwise for flat slabs 8.5/9.5 = 0.89 as span > 8.5 m. * See Appendix B1.5 † In line with Note 5 to Table NA.5, 1.50 is considered to be a maximum for 310/s s s . Cl. 7.4.2(2) This publication is available for purchase from www.concretecentre.com 75 34 l|et s|e| Actual l/d = 9500/260 = 36.5 = OK ‡ Use H20 @ 200 B1 (1570) § c) Flexure: column strip, hogging M Ed = 222.3 kNm/m K = M Ed /bd 2 f ck = 222.3 × 10 6 /(1000 × 260 2 × 30) = 0.109 z/d = 0.89 z = 0.89 × 260 = 231 mm A s = M Ed /f yd z = 222.3 × 10 6 /(231 × 500/1.15) = 2213 mm 2 /m (r = 0.85%) Try H20 @ 125 T1 (2512 mm 2 /m) # Table C5 d) Flexure: middle strip, hogging M Ed = 95.3 kNm/m K = M Ed /bd 2 f ck = 95.3 × 10 6 /(1000 × 260 2 × 30) = 0.069 z/d = 0.95 z = 0.95 × 260 = 247 mm A s = M Ed /f yd z = 95.3 × 10 6 /(247 × 500/1.15) = 887 mm 2 /m (r = 0.34%) Try H16 @ 200 T1 (1005 mm 2 /m) Table C5 e) Requirements i) In column strip, inside middle 1500 mm There is a requirement to place 50% of A t within a width equal to 0.125 of the panel width on either side of the column. Cl. 9.4.1(2) Area required = (3 × 2213 + 3 × 887)/2 mm 2 = 4650 mm 2 Over width = 2 × 0.125 × 6.0 m = 1500 mm i.e. require 4650/1.5 = 3100 mm 2 /m for 750 mm either side of the column centreline. Use H20 @ 100 T1 (3140 mm 2 /m) 750 mm either side of centre of support (16 no. bars) (r = 0.60%) ii) In column strip, outside middle 1500 mm Area required = 3.0 × 2213 – 16 × 314 mm 2 = 1615 mm 2 Over width = 3000 – 2 × 750 mm = 1500 mm i.e. 1077 mm 2 /m Use H20 @ 250 T1 (1256 mm 2 /m) in remainder of column strip ‡ Note: Continuity into columns will reduce sagging moments and criticality of deflection check (see Figures 3.26 and 3.27). § Note requirement for at least 2 bars in bottom layer to carry through column. # The hogging moment could have been considered at face of support to reduce the amount of reinforcement required. Cl. 9.4.1(3) This publication is available for purchase from www.concretecentre.com 76 iii) In middle strip Use H16 @ 200 T1 (1005 mm 2 /m) iv) Perpendicular to edge of slab at edge column Design transfer moment to column M t = 0.17 b e d 2 f ck where Cl. 9.4.2(1), I.1.2(5) b e = c z + y = 400 + 400 = 800 mm M t = 0.17 × 800 × 260 2 × 30 × 10 −6 = 275.8 kNm K = M Ed /bd 2 f ck = 275.8 × 10 6 /(800 × 260 2 × 30) = 0.170 z/d = 0.82 z = 0.82 × 260 = 213 mm A s = M Ed /f yd z = 275.8 × 10 6 /(213 × 500/1.15) = 2978 mm 2 /m This reinforcement to be placed within c x + 2c y = 1100 mm Fig. 9.9 SMDSC [9] Try 10 no. H20 T1 U-bars in pairs @ 200 (3140 mm 2 ) local to column (max. 200 mm from column) Note: Where a 200 × 200 hole occurs on face of column, b e becomes 600 mm and pro rata, A s,req becomes 2233 mm 2 i.e. use 4 no. H20 each side of hole (2512 mm 2 ). v) Perpendicular to edge of slab generally Assuming that there is partial fixity along the edge of the slab, top reinforcement capable of resisting 25% of the moment in the adjacent span should be provided 0.25 × 2213 = 553 mm 2 /m OK Cl. 9.3.1.2(2), 9.2.1.4(1) & NA vi) Check minimum area of reinforcement A s,min = 0.26 (f ctm /f yk ) b t d ≥ 0.0013 b t d where b t = width of tension zone f ctm = 0.30 × f ck 0.666 A s,min = 0.26 × 0.30 × 30 0.666 × 1000 × 260/500 = 390 mm 2 /m (r = 0.15%) Use H12 @ 200 (565 mm 2 /m) Cl. 9.3.1.1, 9.2.1.1 Table 3.1 The reinforcement should extend 0.2h from edge = 600 mm Cl. 9.3.1.4(2) 3.4.6 Analysis grid line 1 (grid 3 similar) Consider grid line 1 as being 9.6/2 + 0.4/2 = 5.0 m wide with continuous spans of 6.0 m. Column strip is 6.0/4 + 0.4/2 = 1.7 m wide. Consider perimeter load is carried by column strip only. Cl. 5.1.1(4) This publication is available for purchase from www.concretecentre.com 77 34 l|et s|e| q k slab = 20.0 kN/m g k cladding = 10.0 kN/m g k slab = 42.5 kN/m 6000 6000 6000 F E D C p Actions: Permanent from slab g k = 5 × 8.5 kN/m 2 = 42.5 kN/m Variable from slab q k = 5 × 4.0 kN/m 2 = 20.0 kN/m Permanent perimeter load g k = 10.0 kN/m Load combination and arrangement: As before, choose to use all-spans-loaded case and coefficients Cl. 5.1.3(1) & NA: Table NA.1 (option c) Ultimate load, n: By inspection, Exp. (6.10b) is critical. Fig. 2.5 n = 1.25 × (42.5 +10) + 1.5 × 20 = 95.6 kN/m EC0: Exp. (6.10b) Perimeter load, 10 × 1.25 = 12.5 kN/m Effective span, l eff Effective span = 6000 – 2 × 400/2 + 2 × 300/2 = 5900 Cl. 5.3.2.2(1) Design moments in bay, M Ed : In spans (worst case, end span assuming pinned support) M Ed = 0.086 × 83.0 × 5.9 2 = 248.5 kNm Table C2 At supports (worst case 1st support) M Ed = 0.086 × 83.0 × 5.9 2 = 248.5 kNm Additional moment in column strip only due to perimeter load, spans (and supports, worst case) M Ed = 0.086 × 12.5 × 5.9 2 = 37.4 kNm Table C2 Apportionment to column strips and middle strips: NA.3 [1a] : Fig. I.1 Apportionment (as %) Column strip, 1.7 m wide Middle strip –ve (hogging) Short span = 75% Short span = 25% +ve (sagging) 50% 50% Table I.1 CS Flat slab guide [27] Short span moments: M Ed Column strip, 1.7 m wide Middle strip, 3.3 m wide –ve (hogging) (0.75 × 248.5 + 37.4)/1.70 = 131.6 kNm/m 0.25 × 248.5/3.3 = 18.8 kNm/m +ve (sagging) (0.50 × 248.5 + 37.4)/1.70 = 95.1 kNm/m 0.50 × 248.5/3.3 = 37.6 kNm/m This publication is available for purchase from www.concretecentre.com 78 Punching shear force, V Ed For internal supports, as before = 516.5 kN For penultimate support, 516.5 × 1.18 = 609.5 kN Table C3 3.4.7 Design grid line 1 (grid 3 similar) Cover: c nom = 30 mm as before d = 300 − 30 − 20 − 20/2 = 240 mm a) Flexure: column strip, sagging M Ed = 95.1 kNm/m K = M Ed /bd 2 f ck = 95.1 × 10 6 /(1000 × 240 2 × 30) = 0.055 Table C5 z/d = 0.95 z = 0.95 × 240 = 228 mm A s = M Ed /f yd z = 95.1 × 10 6 /(228 × 500/1.15) = 959 mm 2 /m (r = 0.40%) Try H16 @ 200 B2 (1005 mm 2 /m) b) Deflection: column strip Check span-to-effective-depth ratio. Appendix B Allowable l/d = N × K × F1 × F2 × F3 where Appendix C7 Tables C10–C13 N = 26.2 (r = 0.40%, f ck = 30) Cl. 7.4.2, Exp. (7.17), Table 7.4N & NA, Table NA.5: Note 5 K = 1.2 (flat slab) F1 = 1.0 (b eff /b w = 1.0) F2 = 1.0 (no brittle partitions) F3 = 310/ s s ≤ 1.5 where s s = s su (A s,req /A s,prov ) 1/d where s su ≈ 283 MPa (from Figure C3 and G k /Q k = 3.6, c 2 = 0.3, g G = 1.25) d = redistribution ratio = 1.08 Fig. C3 = s s ≈ 283 × (959/1005)/1.08 = 250 Table C14 = F3 = 310/250 = 1.24 Fig. C3 =Allowable l/d = 26.2 × 1.2 × 1.24 = 39.0 Actual l/d = 5900/240 = 24.5 = OK Use H16 @ 200 B2 (1005 mm 2 /m) c) Flexure: middle strip, sagging M Ed = 37.6 kNm/m By inspection, z = 228 mm A s = M Ed /f yd z = 37.6 × 10 6 /(228 × 500/1.15) = 379 mm 2 /m (r = 0.56%) This publication is available for purchase from www.concretecentre.com 79 34 l|et s|e| By inspection, deflection OK. Check minimum area of reinforcement. A s,min = 0.26 (f ctm /f yk ) b t d ≥ 0.0013 b t d Cl. 9.3.1.1, 9.2.1.1 where b t = width of tension zone f ctm = 0.30 × f ck 0.666 Table 3.1 A s,min = 0.26 × 0.30 × 30 0.666 × 1000 × 240/500 = 361 mm 2 /m (r = 0.15%) Use H12 @ 300 T2 (376 mm 2 /m) d) Flexure: column strip, hogging M Ed = 131.6 kNm/m K = M Ed /bd 2 f ck = 131.6 × 10 6 /(1000 × 240 2 × 30) = 0.076 z/d = 0.928 z = 0.928 × 240 = 223 mm A s = M Ed /f yd z = 131.6 × 10 6 /(223 × 500/1.15) = 1357 mm 2 /m (r = 0.56%) Try H20 @ 200 T2 (1570 mm 2 /m) ‡ Table C5 e) Flexure: middle strip, hogging M Ed = 18.8 kNm/m By inspection, z = 228 mm Table C5 A s = M Ed /f yd z = 18.8 × 10 6 /(228 × 500/1.15) = 190 mm 2 /m (r = 0.08%) A s,min as before = 361 mm 2 /m (r = 0.15%) Try H12 @ 300 T2 (376 mm 2 /m) Cl. 9.3.1.1, 9.2.1.1 f) Requirements There is a requirement to place 50% of A t within a width equal to 0.125 of the panel width on either side of the column. As this column strip is adjacent to the edge of the slab, consider one side only: Area required = (1.5 × 1357 + 3.3 × 190)/2 mm 2 = 1334 mm 2 Within = 0.125 × 6.0 m = 750 mm of the column centreline, i.e. require 1334/0.75 = 1779 mm 2 /m for 750 mm from the column centreline. Cl. 9.4.1(2) ‡ The hogging moment could have been considered at face of support to reduce the amount of reinforcement required. This should be balanced against the effect of the presence of a 200 × 200 hole at some supports which would have the effect of increasing K but not unduly increasing the total amount of reinforcement required in the column strip (a 1.5% increase in total area would been required). This publication is available for purchase from www.concretecentre.com 80 Allowing for similar from centreline of column to edge of slab: Use 6 no. H20 @ 175 T2(1794 mm 2 /m) (r = 0.68%) between edge and to 750 mm from centre of support In column strip, outside middle 1500 mm, requirement is for 1.7 × 1357 – 6 × 314 = 422 mm 2 in 750 mm, i.e. 563 mm 2 /m Use H12 @ 175 T2 (646 mm 2 /m) in remainder of column strip In middle strip Use H12 @ 300 T2 (376 mm 2 /m) 3.4.8 Analysis grid line 2 Consider panel on grid line 2 as being 9.6/2 + 8.6/2 = 9.1 m wide and continuous spans of 6.0 m. Column strip is 6.0/2 = 3.0 m wide. (See Figure 3.20). q k slab = 36.4 kN/m g k slab = 77.4 kN/m 6000 6000 6000 F E D C Figure 3.22 Internal panel on grid 2 p Slab g k = 9.1 × 8.5 kN/m 2 = 77.4 kN/m Slab q k = 9.1 × 4.0 kN/m 2 = 36.4 kN/m Actions, load combination and arrangement: Choose to use all-spans-loaded case. Cl. 5.1.3(1) & NA: Table NA.1 (option c) Ultimate load, n: By inspection, Exp. (6.10b) is critical. n = 1.25 × 77.4 + 1.5 × 36.4 = 151.4 kN/m Fig. 2.5 EC0: Exp. (6.10b) Effective span, l eff : Effective span = 5900 mm as before. Cl. 5.3.2.2(1) Design moments in bay, M Ed : Spans (worst case, end span assuming pinned support) M Ed = 0.086 × 151.4 × 5.9 2 = 453.2 kNm Table C2 Support (worst case 1st support) M Ed = 0.086 × 151.4 × 5.9 2 = 453.2 kNm Additional moment in column strip only due to perimeter load. Table C2 This publication is available for purchase from www.concretecentre.com 81 34 l|et s|e| Apportionment to column strips and middle strips: M Ed Column strip, 3.0 m wide Middle strip, 6.1 m wide –ve (hogging) 0.75 × 453.2/3.0 = 113.3 kNm/m 0.25 × 453.2/6.1 = 18.5 kNm/m +ve (sagging) 0.50 × 453.2/3.0 = 75.5 kNm/m 0.50 × 453.2/6.1 = 37.1 kNm/m Punching shear force, V Ed , as before. 3.4.9 Design grid line 2 Effective depth, d d = 300 − 30 − 20 − 20/2 = 240 mm a) Flexure: column strip, sagging M Ed = 75.5 kNm/m By inspection, z = 228 mm A s = M Ed /f yd z = 75.5 × 10 6 /(228 × 500/1.15) = 761 mm 2 /m (r = 0.32%) Try H16 @ 250 B2 (804 mm 2 /m) Table C5 Deflection: column strip By inspection, OK. b) Flexure: column strip, sagging M Ed = 37.1 kNm/m By inspection, z = 228 mm A s = M Ed /f yd z = 37.1 × 10 6 /(228 × 500/1.15) = 374 mm 2 /m (r = 0.55%) By inspection, deflection OK. Try H10 @ 200 B2 (393 mm 2 /m) c) Flexure: column strip, hogging M Ed = 113.3 kNm/m K = M Ed /bd 2 f ck = 113.3 × 10 6 /(1000 × 240 2 × 30) = 0.065 z/d = 0.94 z = 0.928 × 240 = 225 mm Table C5 A s = M Ed /f yd z = 113.3 × 10 6 /(225 × 500/1.15) = 1158 mm 2 /m (r = 0.48%) Try H20 @ 250 T2 (1256 mm 2 /m) ‡ d) Flexure: middle strip, hogging M Ed = 18.5 kNm/m By inspection, z = 228 mm ‡ The hogging moment could have been considered at face of support to reduce the amount of reinforcement required. This publication is available for purchase from www.concretecentre.com 82 A s = M Ed /f yd z = 18.5 × 10 6 /(228 × 500/1.15) = 187 mm 2 /m (r = 0.08%) Table C5 As before minimum area of reinforcement governs A s,min = 0.26 × 0.30 × 30 0.666 × 1000 × 240/500 = 361 mm 2 /m (r = 0.15%) Try H12 @ 300 B2 (376 mm 2 /m) Cl. 9.3.1.1, 9.2.1.1 e) Requirements Regarding the requirement to place 50% of A t within a width equal to 0.125 of the panel width on either side of the column: Area required = (3.0 × 1158 + 6.1 × 187)/2 mm 2 = 2307 mm 2 Within = 2 × 0.125 × 6.0 m = 1500 mm centred on the column centreline, i.e. require 2307/1.5 = 1538 mm 2 /m for 750 mm either side of the column centreline. Use H20 @ 200T2 (1570 mm 2 /m) 750 mm either side of centre of support (r = 0.60%) In column strip, outside middle 1500 mm, requirement is for 3.0 × 1158 – 1.5 × 1570 = 1119 mm 2 in 1500 mm, i.e. 764 mm 2 /m Use H16 @ 250 T2 (804 mm 2 /m) in remainder of column strip In middle strip: Use H12 @ 300 T2 (376 mm 2 /m) 3.4.10 Punching shear, central column, C2 At C2, applied shear force, V Ed = 1204.8 kN ‡ a) Check at perimeter of column v Ed = bV Ed /u i d < v Rd,max where Cl. 6.4.3(2), 6.4.5(3) b = factor dealing with eccentricity; recommended value 1.15 V Ed = applied shear force Fig. 6.21N & NA u i = control perimeter under consideration. For punching shear adjacent to interior columns Cl. 6.4.5(3) u 0 = 2(c x + c y ) = 1600 mm d = mean effective depth = (260 + 240)/2 = 250 mm Exp. (6.32) v Ed = 1.15 × 1204.8 × 10 3 /1600 × 250 = 3.46 MPa v Rd,max = 0.5vf cd Cl. 6.4.5(3) Note ‡ Column C2 is taken to be an internal column. In the case of a penultimate column, an additional elastic reaction factor should have been considered. This publication is available for purchase from www.concretecentre.com 83 34 l|et s|e| where v = 0.6(1 − f ck /250) = 0.528 f cd = a cc lf ck /g C = 1.0 × 1.0 × 30/1.5 = 20 = 0.5 × 0.528 × 20 = 5.28 MPa = OK Table C7 § b) Check shear stress at control perimeter u 1 (2d from face of column) v Ed = bV Ed /u 1 d < v Rd,c where b, V Ed and d as before Cl. 6.4.2 u 1 = control perimeter under consideration. For punching shear at 2d from interior columns u 1 = 2(c x + c y ) + 2Q × 2d = 4741 mm Fig. 6.13 v Ed = 1.15 × 1204.8 × 10 3 /4741 × 250 = 1.17 MPa v Rd,c = 0.18/ g C k (100 r l f ck ) 0.333 Exp. (6.47) & NA where g C = 1.5 k = 1 + (200/d) 0.5 ≤ 2 k = 1 + (200/250) 0.5 = 1.89 r l = (r ly r lz )0.5 = (0.0085 × 0.0048) 0.5 = 0.0064 Cl. 6.4.4.1(1) where r |y , r lz = Reinforcement ratio of bonded steel in the y and z direction in a width of the column plus 3d each side of column # f ck = 30 v Rd,c = 0.18/1.5 × 1.89 × (100 × 0.0064 × 30) 0.333 = 0.61 MPa = Punching shear reinforcement required Table C5* c) Perimeter at which punching shear links are no longer required u out = V Ed × b/(d v Rd,c ) u out = 1204.8 × 1.15 × 10 3 /(250 × 0.61) = 9085 mm Exp. (6.54) Length of column faces = 4 × 400 = 1600 mm Radius to u out = (9085 – 1600)/2Q = 1191 mm from face of column Perimeters of shear reinforcement may stop 1191 – 1.5 × 250 = 816 m from face of column Cl. 6.4.5(4) & NA Shear reinforcement (assuming rectangular arrangement of links): s r,max = 250 × 0.75 = 187, say = 175 mm Cl. 9.4.3(1) § At the perimeter of the column, v Rd,max assumes the strut angle is 45°, i.e. that cot y = 1.0. Where cot y = < 1.0, v Rd,max is available from Table C7. # The values used here for r ly , r lz ignore the fact that the reinforcement is concentrated over the support. Considering the concentration would have given a higher value of V Rd,c at the expense of further calculation to determine r ly , r lz at 3d from the side of the column. * v Rd,c for various values of d and r l is available from Table C6. This publication is available for purchase from www.concretecentre.com 84 Inside 2d control perimeter, s t,max = 250 × 1.5 = 375, say 350 mm Cl. 9.4.3(2) Outside control perimeter s t,max = 250 × 2.0 = 500 mm Assuming vertical reinforcement: At the basic control perimeter, u 1 , 2d from the column ‡ : A sw ≥ (v Ed – 0.75v Rd,c ) s r u 1 /1.5f ywd,ef ) Exp. (6.52) where f ywd,ef = effective design strength of reinforcement = (250 + 0.25d) < f yd = 312 MPa Cl. 6.4.5(1) For perimeter u 1 A sw = (1.17 – 0.75 × 0.61) × 175 × 4741/(1.5 × 312) = 1263 mm 2 per perimeter A sw,min ≥ 0.08f ck 0.5 (s r s t )/(1.5 f yk sin a + cos a) Exp. (9.11) where A sw,min = minimum area of a single leg of link a = angle between main reinforcement and shear reinforcement; for vertical reinforcement sin a = 1.0 A sw,min ≥ 0.08 × 30 0.5 (175 × 350)/(1.5 × 500) = 36 mm 2 = Try H8 legs of links (50 mm 2 ) A sw /u 1 ≥ 1263/4741 = 0.266 mm 2 /mm Using H8 max. spacing = min[50/0.266; 1.5d] = min[188; 375] = 188 mm cc = Use min. H8 legs of links at 175 mm cc around perimeter u 1 Cl. 9.4.3 Perimeters at 0.75d = 0.75 × 250 = 187.5 mm say = 175 mm centres Cl. 9.4.3(1) d) Check area of reinforcement > 1263 mm 2 in perimeters inside u 1 § 1st perimeter to be > 0.3d but < 0.5d from face of column. Say 0.4d = 100 mm from face of column. By inspection of Figure 3.23 the equivalent of 10 locations are available at 0.4d from column therefore try 2 × 10 no. H10 = 1570 mm 2 . Fig. 9.10, Cl. 9.4.3(4) By inspection of Figure 3.23 the equivalent of 18 locations are available at 1.15d from column therefore try 18 no. H10 = 1413 mm 2 . By inspection of Figure 3.23 the equivalent of 20 locations are available at 1.90d from column therefore try 20 no. H10 = 1570 mm 2 . By inspection of Figure 3.23 beyond u 1 to u out grid of H10 at 175 x 175 OK. ‡ Clause 6.4.5 provides Exp. (6.52), which by substituting v Ed for v Rd,c , allows calculation of the area of required shear reinforcement, A sw , for the basic control perimeter, u 1 . § The same area of shear reinforcement is required for all perimeters inside or outside perimeter u 1 . See Commentary on design, Section 3.4.14. Punching shear reinforcement is also subject to requirements for minimum reinforcement and spacing of shear reinforcement (see Cl. 9.4.3). Cl. 6.4.5 Exp. 6.5.2 Cl. 9.4.3 This publication is available for purchase from www.concretecentre.com 85 34 l|et s|e| e) Summary of punching shear refreshment required at column C2 175 u out Punching shear reinforcement no longer required 1.5d = 375 S = 112 H10 legs of links u 1 at 2d from column u out 1 7 5 1 7 5 1 7 5 1 7 5 1 7 5 1 7 5 1 7 5 1 7 5 2 0 0 2 0 0 2 0 0 375 716 7 1 6 7 1 6 1 0 0 1 0 0 4 0 0 716 100 400 100 175 175 175 175 175 175 175 200 200 200 C 2 Figure 3.23 Punching shear links at column C2 (112 no. links) (column D2 similar) 3.4.11 Punching shear, edge column Assuming penultimate support, V Ed = 1.18 × 516.5 = 609.5 kN Table C3 a) Check at perimeter of column v Ed = bV Ed /u i d < v Rd,max where Cl. 6.4.3(2), 6.4.5(3) b = factor dealing with eccentricity; recommended value 1.4 V Ed = applied shear force Fig. 6.21N & NA u i = control perimeter under consideration. For punching shear adjacent to edge columns u 0 = c 2 + 3d < c 2 + 2c 1 = 400 + 750 < 3 × 400 mm = 1150 mm Cl. 6.4.5(3) d = as before 250 mm v Ed = 1.4 × 609.5 × 10 3 /1150 × 250 = 2.97 MPa Exp. (6.32) v Rd,max , as before = 5.28 MPa = OK Cl. 6.4.5(3) Note This publication is available for purchase from www.concretecentre.com 86 b) Check shear stress at basic perimeter u 1 (2.0d from face of column) Cl. 6.4.2 v Ed = bV Ed /u 1 d < v Rd,c where b, V Ed and d as before u 1 = control perimeter under consideration. For punching shear at 2d from edge column columns u 1 = c 2 + 2c 1 + Q × 2d = 2771 mm v Ed = 1.4 × 609.5 × 10 3 /2771 × 250 = 1.23 MPa Fig. 6.15 v Rd,c = 0.18/ g C × k × (100 r l f ck ) 0.333 where Exp. (6.47) & NA g C = 1.5 k = as before = 1 +(200/250) 0.5 = 1.89 r l = (r ly r lz ) 0.5 where r |y , r lz = Reinforcement ratio of bonded steel in the y and z direction in a width of the column plus 3d each side of column. r ly : (perpendicular to edge) 10 no. H20 T2 + 6 no. H12 T2 in 2 × 750 + 400, i.e. 3818 mm 2 in 1900 mm =r ly = 3818/(250 × 1900) = 0.0080 r lz : (parallel to edge) 6 no. H20 T1 + 1 no. T12 T1 in 400 + 750 i.e. 1997 mm 2 in 1150 mm. =r lz = 1997/(250 × 1150) = 0.0069 r l = (0.0080 × 0.0069) 0.5 = 0.0074 f ck = 30 Cl. 6.4.4.1(1) v Rd,c = 0.18/1.5 × 1.89 × (100 × 0.0074 × 30) 0.333 = 0.64 MPa = Punching shear reinforcement required Table C6 ‡ 3d = 750 H12 @ 200 U-bars H12 @ 175T1 6H20T1 @175 10H20 U-bars in pairs @ 200 cc 3d = 750 400 4 0 0 3 d = 7 5 0 C 1 Figure 3.24 Flexural tensile reinforcement adjacent to columns C1 (and C3) ‡ v Rd,c for various values of d and r l is available from Table C6. This publication is available for purchase from www.concretecentre.com 87 34 l|et s|e| c) Perimeter at which punching shear links no longer required u out = 609.5 × 1.4 × 10 3 /(250 × 0.64) = 5333 mm Length attributable to column faces = 3 × 400 = 1200 mm = radius to u out from face of column = say (5333 − 1200)/Q = 1315 mm from face of column Exp. (6.54) Perimeters of shear reinforcement may stop 1370 – 1.5 × 250 = 940 mm from face of column. Cl. 6.4.5(4) & NA d) Shear reinforcement As before, s r,max = 175 mm; s t,max = 350 mm and f ywd,ef = 312 MPa For perimeter u 1 Cl. 9.4.3(1), 9.4.3(2) A sw ≥ (v Ed – 0.75v Rd,c ) s r u 1 /1.5f ywd,ef = (1.23 – 0.75 × 0.64) × 175 × 2771/(1.5 × 312) = 777 mm 2 per perimeter Exp. (6.52) A sw,min ≥ 0.08 × 30 0.5 (175 × 350)/(1.5 × 500) = 36 mm 2 A sw /u 1 ≥ 777/2771 = 0.28 mm 2 /mm Using H8 max. spacing = 50/0.28 = 178 mm cc =Use min. H8 (50 mm 2 ) legs of links at 175 mm cc around perimeters: perimeters at 175 mm centres Exp. (9.11) e) Check area of reinforcement > 777 mm 2 in perimeters inside u 1 § 1st perimeter to be > 0.3d but < 0.5d from face of column. Say 0.4d = 100 mm from face of column Fig. 9.10, Cl. 9.4.3(4) By inspection of Figure 3.27 the equivalent of 6 locations are available at 0.4d from column therefore try 2 × 6 no. H10 = 942 mm 2 By inspection of Figure 3.27 the equivalent of 12 locations are available at 1.15d from column therefore try 12 no. H10 = 942 mm 2 By inspection of Figure 3.27 the equivalent of 14 locations are available at 1.90d from column therefore try 14 no. H10 = 1099 mm 2 By inspection of Figure 3.27 beyond u 1 to u out grid of H10 at 175 x 175 OK. 3.4.12 Punching shear, edge column with hole Check columns D1 and D3 for 200 × 200 mm hole adjacent to column. As previously described use 4 no. H20 U-bars each side of column for transfer moment. Assuming internal support, V Ed = 516.5 kN § See Commentary on design Section 3.4.14. Punching shear reinforcement is also subject to requirements for minimum reinforcement and spacing of shear reinforcement (see Cl. 9.4.3). Cl. 9.4.3 This publication is available for purchase from www.concretecentre.com 88 a) Check at perimeter of column v Ed = bV Ed /u i d < v Rd,max where Cl. 6.4.3(2), 6.4.5(3) b = factor dealing with eccentricity; recommended value 1.4 V Ed = applied shear force Fig. 6.21N & NA u i = control perimeter under consideration. For punching shear adjacent to edge columns u 0 = c 2 + 3d < c 2 + 2c 1 = 400 + 750 < 3 × 400 mm = 1150 mm Allowing for hole, u 0 = 1150 – 200 = 950 mm Cl. 6.4.5(3) d = 250 mm as before Exp. (6.32) v Ed = 1.4 × 516.5 × 10 3 /950 × 250 = 3.06 MPa v Rd,max as before = 5.28 MPa = OK Cl. 6.4.5(3) Note b) Check shear stress at basic perimeter u 1 (2.0d from face of column) Cl. 6.4.2 v Ed = bV Ed /u 1 d < v Rd,c where b, V Ed and d as before u 1 = control perimeter under consideration. For punching shear at 2d from edge column columns Fig. 6.15 u 1 = c 2 + 2c 1 + Q × 2d = 2771 mm Allowing for hole 200/(c 1 /2): x/(c 1 /2 + 2d) 200/200: x/( 200 + 500) = x = 700 mm u 1 = 2771 – 700 = 2071 mm Fig. 6.14 v Ed = 1.4 × 516.5 × 10 3 /2071 × 250 = 1.40 MPa v Rd,c = 0.18/g C × k × (100 r l f ck ) 0.333 Exp. (6.47) & NA where g C = 1.5 k = as before = 1 + (200/250) 0.5 = 1.89 r l = (r l y r l z ) 0.5 where r |y , r lz = Reinforcement ratio of bonded steel in the y and z direction in a width of the column plus 3d each side of column Cl. 6.4.4.1(1) r ly : (perpendicular to edge) 8 no. H20 T2 + 6 no. H12 T2 in 2 × 720 + 400 − 200, i.e. 3190 mm 2 in 1640 mm. =r l y = 3190/(240 × 1640) = 0.0081 r lz : (parallel to edge) 6 no. H20 T1 (5 no. are effective) + 1 no. T12 T1 in 400 + 750 – 200, i.e. 1683 mm 2 in 950 mm. =r l z = 1683/(260 × 950) = 0.0068 This publication is available for purchase from www.concretecentre.com 89 34 l|et s|e| r l = (0.0081 × 0.0068) 0.5 = 0.0074 f ck = 30 v Rd,c = 0.18/1.5 × 1.89 × (100 × 0.0074 × 30) 0.33 = 0.64 MPa Table C6 ‡ = punching shear reinforcement required 3d = 750 H12 @ 200 U-bars H12 @ 175T1 6H20T @ 175 8H20 U-bars in pairs @ 200 cc 3d = 750 400 4 0 0 3 d = 7 5 0 3 D Figure 3.25 Flexural tensile reinforcement adjacent to columns D1 and D3 c) Perimeter at which punching shear links no longer required Exp. (6.54) u out = 516.5 × 1.4 × 10 3 /(250 × 0.64) = 4519 mm Length attributable to column faces = 3 × 400 = 1200 mm Angle subtended by hole from centre of column D1 (See Figures 3.25 & 3.27) = 2 tan −1 (100/200) = 2 × 26.5° = 0.927 rads. = radius to u out from face of column = say (4519 − 1200)/(Q − 0.927) = 1498 mm from face of column Perimeters of shear reinforcement may stop 1498 – 1.5 × 250 = 1123 mm from face of column Cl. 6.4.5(4) & NA d) Shear reinforcement As before, s r,max = 175 mm; s t,max = 350 mm and f ywd,ef = 312 MPa Cl. 9.4.3(1) 9.4.3(2) For perimeter u 1 A sw ≥ (v Ed – 0.75v Rd,c ) s r u 1 /1.5f ywd,ef ) per perimeter = (1.40 – 0.75 × 0.64) × 175 × 2071/(1.5 × 312) = 712 mm 2 per perimeter Exp. (6.52) A sw,min ≥ 0.08 × 30 0.5 (175 × 350)/(1.5 × 500) = 36 mm 2 A sw /u 1 ≥ 712/2071 = 0.34 mm 2 /mm ‡ v Rd,c for various values of d and r l is available from Table C6. This publication is available for purchase from www.concretecentre.com 90 Using H8 (50 mm 2 ) max. spacing = min[50/0.3; 1.5d] = min[147; 375] = 147 mm cc No good Try using H10, max. spacing = 78.5/0.34 = 231 mm cc, say 175 cc = Use min. H10 (78.5 mm 2 ) legs of links at 175 mm cc around perimeters: perimeters at 175 mm centres Check min. 9 no. H10 legs of links (712 mm 2 ) in perimeter u 1 , 2 d from column face. e) Check area of reinforcement > 712 mm 2 in perimeters inside u 1 ‡ 1st perimeter to be 100 mm from face of column as before. Fig. 9.10, Cl. 9.4.3(4) By inspection of Figure 3.27 the equivalent of 6 locations are available at 0.4d from column therefore try 2 × 6 no. H10 = 942 mm 2 . By inspection of Figure 3.27 the equivalent of 10 locations are available at 1.15d from column therefore try 10 H10 = 785 mm 2 . By inspection of Figure 3.27 beyond 1.15d to u out grid: H10 at 175 x 175 OK. 3.4.13 Summary of design Grid C flexure End supports: Column strip: (max. 200 mm from column) 10 no. H20 U-bars in pairs (where 200 × 200 hole use 8 no. H20 T1 in U-bars in pairs) Middle strip: H12 @ 200 T1 Spans 1–2 and 2–3: Column strip and middle strip: H20 @ 200 B Central support: Column strip centre: for 750 mm either side of support: H20 @ 100 T1 Column strip outer: H20 @ 250 T1 Middle strip: H16 @ 200 T1 Grid 1 (and 3) flexure Spans: Column strip: H16 @ 200 B2 Middle strip: H12 @ 300 B2 ‡ See Commentary on design Section 3.4.14. Punching shear reinforcement is also subject to requirements for minimum reinforcement and spacing of shear reinforcement. Cl. 9.4.3 This publication is available for purchase from www.concretecentre.com 91 34 l|et s|e| Interior support: Column strip centre: 6 no. H20 @ 175 T2 Column strip outer: H12 @ 175 T2 Middle strip: H12 @ 300 T2 Grid 2 flexure Spans: Column strip: H16 @ 250 B2 Middle strip: H10 @ 200 B2 Interior support: Column strip centre: H20 @ 200 T2 Column strip outer: H16 @ 250 T2 Middle strip: H12 @ 300 T2 See Figure 3.26 Punching shear Internal (e.g. at C2): Generally, use H10 legs of links in perimeters at max. 175 mm centres, but double up on 1st perimeter Max. tangential spacing of legs of links, s t,max = 270 mm Last perimeter, from column face, min. 767 mm See Figure 3.26 Edge (e.g. at C1, C3 assuming no holes): Generally, use H10 legs of links in perimeters at max. 175 mm centres but double up on 1st perimeter Max. tangential spacing of legs of links, s t,max = 175 mm Last perimeter, from column face, min. 940 mm Edge (e.g. at D1, D3 assuming 200 × 200 hole on face of column): Generally, use H10 legs of links in perimeters at max. 175 mm centres but double up on 1st perimeter Max. tangential spacing of legs of links, s t,max = 175 mm Last perimeter, from column face, min. 1123 mm See Figure 3.27 This publication is available for purchase from www.concretecentre.com 92 Commentary on design a) Method of analysis The use of coefficients in the analysis would not usually be advocated in the design of such a slab. Nonetheless, coefficients may be used and, unsurprisingly, their use leads to higher design moments and shears, as shown below. Method Moment in 9.6 m span per 6 m bay (kNm) Centre support moment per 6 m bay (kNm) Centre support reaction V Ed (kN) Coefficients 842.7 952.8 1205 Continuous beam 747.0 885.6 1103 Plane frame columns below 664.8 834.0 1060 Plane frame columns above and below 616.8 798.0 1031 3.4.14 10H20 T1 U-bars in pairs @ 200 6H20 - 175 T2 H12 - 200 T1 U-bars 8H20 T1 U-bars in pairs @ 200 H16 @ 175 B2* 5H12 @ 175 T2 H12 @ 300 B2 H10 @ 200 T2 H16 @ 175 B2* 9H20 @ 175 T2* 3H20 @ 250 T1 H20 @ 200 B1 H12 @ 200 T1 U-bars 16H20 @ 100 T1 3H20 @ 250 T1 H16 @ 200 T1 3H20 @ 250 T1 16H20 @ 100 T1 4H16 @ 250 T2 C D 1 2 Note:* Spacing rationalised to suit punching shear links Figure 3.26 Reinforcement details bay C–D, 1–2 This publication is available for purchase from www.concretecentre.com 93 34 l|et s|e| u out u out u 1 1123 500 8H20 T1 U-bars in pairs H10 @ 200 T1 U-bars 6H20 @ 175 T2 1.5d 6H16 @ 175 B2 H10 @ 200 T1 U-bars 375 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 175 200 100 100 C L 200 S = 152 H10 legs of links @ 175 mm centres Ineffective area 1 1 D Note: For internal column see Figure 3.23 Figure 3.27 Punching shear links at column D1 (and D3) (penultimate support without hole similar) These higher moments and shears result in rather more reinforcement than when using other more refined methods. For instance, the finite element analysis used in Guide to the design and construction of reinforced concrete flat slabs [27] for this bay, leads to: H16 @ 200 B1 in spans 1–2 (cf. H20 @ 200 B1 using coefficients) t H20 @ 125 T1 at support 2 (cf. H20 @ 100 T1 using coefficients) t 3 perimeters of shear links at C2 for t V Ed = 1065 kN (cf. 5 perimeters using coefficients) 2 perimeters of shear links at C3 (cf. 7 perimeters using coefficients) t b) Effective spans and face of support In the analysis using coefficients, advantage was taken of using effective spans to calculate design moments. This had the effect of reducing span moments. Cl. 5.3.2.2(1) At supports, one may base the design on the moment at the face of support. This is borne out by Guide to the design and construction of reinforced concrete flat slabs [27] that states that hogging moments greater than those at a distance h c /3 may be ignored (where h c is the effective diameter of a column or column head). This is in line with BS 8110 [7] and could have been used to reduce support moments. Cl. 5.3.2.2(3) This publication is available for purchase from www.concretecentre.com 94 c) Punching shear reinforcement Arrangement of punching shear links According to the literal definition of / sw in Exp. (6.52), the same area of shear reinforcement is required for all perimeters inside or outside perimeter a ! (rather than (/ sw /a ! )/s r being considered as the required density of shear reinforcement on and within perimeter a ! ). For perimeters inside a ! , it might be argued that Exp. (6.50) Exp. 6.52 Exp. 6.50 (enhancement close to supports) should apply. However, at the time of writing, this expression is deemed applicable only to foundation bases. Therefore, large concentrations of shear reinforcement are required close to the columns – in this example, this included doubling up shear links at the 1st perimeter. Similar to BS 8110 [7] figure 3.17, it is apparent that the requirement for punching shear reinforcement is for a punching shear zone 1.5d wide. However, in Eurocode 2, the requirement has been ‘simplified’ in Exp. (6.52) to make the requirement for a perimeter (up to 0.75d wide). It might appear reasonable to apply the same 40%:60% rule (BS 8110 Cl. 3.7.7.6) to the first two perimeters to make doubling of punching shear reinforcement at the first perimeter unnecessary: in terms of Eurocode 2 this would mean 80% A sw on the first perimeter and 120% A sw on the second. Using this arrangement it would be possible to replace the designed H10 links in the first two perimeters with single H12 links. BS 8110: Fig. 3.17 BS 8110: Cl. 3.7.7.6 Outside u 1 , the numbers of links could have been reduced to maintain provision of the designed amount of reinforcement A sw . A rectangular arrangement of H12 links would have been possible (within perimeter u 1 , 350 × 175; outside u 1 , 500 × 175). However, as the grid would need to change orientation around each column (to maintain the 0.75d radial spacing) and as the reinforcement in B2 and T2 is essentially at 175 centres, it is considered better to leave the arrangement as a regular square grid. Cl. 9.4.3(1) Use of shear reinforcement in a radial arrangement, e.g. using stud rails, would have simplified the shear reinforcement requirements. V Ed /V Rd,c In late 2008, a proposal was made for the UK National Annex to include a limit of 2.0 or 2.5 on V Ed /V Rd,c (or v Ed /v Rd,c ) within punching shear requirements. It is apparent that this limitation could have major effects on flat slabs supported on relatively small columns. For instance in Section 3.4.12, edge column with hole, V Ed /V Rd,c = 2.18. Curtailment of reinforcement In this design, the reinforcement would be curtailed and this would be done either in line with previous examples or, more practically, in line with other guidance [20, 21] . This publication is available for purchase from www.concretecentre.com 95 3S Ste|r ¦||ght lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Stair flight 600 600 9 @250 3450 1750 Figure 3.28 Stair flight 3.5.1 Loads kN/m 2 Permanent (worse case flight) Assume 160 waist 0.160 × 305/250 × 25 = 4.88 Treads 4 × 0.25 × 0.175/2 × 25 = 2.19 50 mm screed 0.5 × 22 = 1.10 Finishing = 0.03 g k = 8.20 Variable action: crowd loading q k = 4.00 EC1-1-1: Table 6.1, 6.2 & NA.3 3.5.2 Moment M Ed = (8.20 × 1.25 + 4.00 × 1.5) × 3.45 2 /8 = 24.2 kNm/m 3.5.3 Design d = 160 – c nom – f/2 where c nom = 25 mm (for XC1) f = 12 mm (assumed) Concise: Table 4.2; BS 8500 = d = 129 mm Stair flight ¹h|s exemp|e |s ¦or e typ|ce| ste|r ¦||ght 3.5 This publication is available for purchase from www.concretecentre.com 96 K = M Ed /bd 2 f ck = 24.2 × 10 6 /(1000 × 129 2 × 30) = 0.048 z/d = 0.95 Table C5 z = 0.95 × 129 = 122 mm A s = M Ed /f yd z = 24.2 × 10 6 /[(500/1.15) × 122] = 456 mm 2 /m (r = 0.35%) Try H12 @ 250 (452 mm 2 /m) = OK) 3.5.4 Check deflection Allowable l/d = N × K × F1 × F2 × F3 Appendix C7, Table C10 where N = 32.7 K = 1.0 Table C11 F1 = 1.0 F2 = 1.0 F3 = 1.0 (say) = Allowable l/d = 32.7 Actual l/d = 3450/129 = 26.7 = OK =Provide H12 @ 250 B. This publication is available for purchase from www.concretecentre.com 4! Cont|nuous |eem on p|n supports 97 Beams General ¹he ce|cu|et|ons |n th|s Sect|on ere presented |n the ¦o||ow|ng perts 4! Cont|nuous |eem on p|n supports ÷ e s|mp|y supported cont|nuous |eem show|ng whet m|ght |e deemed typ|ce| hend ce|cu|et|ons 4? A heev||y |oeded l-|eem 43 A cont|nuous w|de ¹-|eem ¹h|s exemp|e |s ene|ysed end des|gned str|ct|y |n eccordence w|th the prov|s|ons o¦ lurocode ? ¹hey ere |ntended to |e |||ustret|ve o¦ the Code end not necesser||y |est prect|ce A genere| method o¦ des|gn|ng |eems |s shown |e|ow ln prect|ce, severe| o¦ these steps mey |e com||ned Leterm|ne des|gn ||¦e N EC0 & NA Table NA.2.1 Assess ect|ons on the |eem N EC1 & NAs Assess dure||||ty requ|rements end determ|ne concrete N strength Table 4.1 BS 8500–1: Tables A4, A5 Chec| cover requ|rements ¦or eppropr|ete ¦|re N res|stence per|od EC2–1–2: Tables 5.8, 5.9, 5.10, 5.11 Ce|cu|ete m|n|mum cover ¦or dure||||ty, ¦|re end |ond N requ|rements Cl. 4.4.1 Leterm|ne wh|ch com||net|ons o¦ ect|ons epp|y N EC0 & NA Tables NA.A1.1, NA.A1.2 (B) Leterm|ne |oed|ng errengements N Cl. 5.1.3(1) & NA Ane|yse structure to o|te|n cr|t|ce| moments end sheer N ¦orces Cl. 5.4, 5.5, 5.6 Les|gn ¦|exure| re|n¦orcement N Cl. 6.1 Chec| de¦|ect|on N Cl. 7.4 Chec| sheer cepec|ty N Cl. 6.2 Other des|gn chec|s N Chec| m|n|mum re|n¦orcement Chec| crec||ng (s|.e or spec|ng o¦ |ers) Chec| e¦¦ects o¦ pert|e| ux|ty Chec| secondery re|n¦orcement Cl. 9.3.1.1(1), 9.2.1.1(1) Cl. 7.3, Tables 7.2N, 7.3N Cl. 9.3.1.2(2) Cl. 9.3.1.1(2), 9.3.1.4(1) Chec| curte||ment N Cl. 9.3.1.1(4), 9.2.1.3, Fig. 9.2 Chec| enchorege N Cl. 9.3.1.2, 8.4.4, 9.3.1.1(4), 9.2.1.5(1), 9.2.1.5(2) Chec| |eps N Cl. 8.7.3 4 4.0 This publication is available for purchase from www.concretecentre.com 98 4.1 A 450 mm deep × 300 mm wide rectangular beam is required to support office loads of g k = 30.2 kN/m and q k = 11.5 kN/m over 2 no. 6 m spans. f ck f = 30 MPa, f yk f = 500 MPa. Assume 300 mm wide supports, a 50-year design life and a requirement for a 2-hour resistance to fire in an external but sheltered environment. q k = 11.5 kN/m g k = 30.2 kN/m 6000 6000 Figure 4.1 Continuous rectangular beam 450 300 Figure 4.2 Section through beam 4.1.1 Actions Permanent g k = 30.2 kN/m and variable q k = 11.5 kN/m 4.1.2 Cover Nominal cover, c nom : c nom = c min + Dc dev Exp. (4.1) where c min = max[c min,b ; c min,dur ] where c min,b = minimum cover due to bond = diameter of bar. Assume 25 mm main bars c min,dur = minimum cover due to environmental conditions. Assuming XC3 (moderate humidity or cyclic wet and dry) and secondarily XF1 (moderate water Cl. 4.4.1.2(3) lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Continuous beam on pin supports Continuous beam on pin supports ¹h|s ce|cu|et|on |s |ntended to show e typ|ce| hend ce|cu|et|on ¦or e cont|nuous s|mp|y supported |eem us|ng coe¦uc|ents to determ|ne moments end sheers This publication is available for purchase from www.concretecentre.com 4! Cont|nuous |eem on p|n supports 99 saturation without de-icing salt) using C30/37 concrete, c min,dur = 25 mm Table C3 BS 8500-1 [14] : Table A4; How to: Building structures [8] Dc dev = allowance in design for deviation. Assuming no measurement of cover, Dc dev = 10 mm Cl. 4.4.1.2(3) = c nom = 25 + 10 = 35 mm Fire: Check adequacy of section for 2 hours fire resistance (i.e. REI = 120) For b min = 300 mm, minimum axis distance, a = 35 mm = OK c nom = 35 mm EC2-1-2: 5.6.3(1), Table 5.6 4.1.3 Load combination (and arrangement) Load combination: By inspection, BS EN 1990 Exp. (6.10b) governs = n = 1.25 × 30.2 + 1.5 × 11.5 = 50.8 kN/m Fig. 2.5 EC0: Exp. (6.10b) Arrangement: Choose to use all-and-alternate-spans-loaded load cases, i.e. use coefficients. Cl. 5.1.3(1) & NA Table NA.1 (option b) Table C3 The coefficients used assume 15% redistribution at supports. As the amount of redistribution is less than 20%, there are no restrictions on reinforcement grade. The use of Table 5.6 in BS EN 1992–1–2 is restricted to where redistribution does not exceed 15%. Table C3 Cl. 5.5(4) & NA EC2-1-2: 5.6.3(1), Table 5.6 4.1.4 Analysis Design moments: Spans M Ed = (1.25 × 30.2 × 0.090 + 1.5 × 11.5 × 0.100) × 6.0 2 = 122.3 + 62.1 = 184.4 kNm Appendix C1, Table C3 Support M Ed = 50.8 × 0.106 × 6.0 2 = 193.8 kNm Table C3 Shear force: V AB = 0.45 × 6.0 × 50.8 = 137.2 kN V AB = 0.63 × 6.0 × 50.8 = 192.0 kN 4.1.5 Flexural design Effective depth: Assuming 10 mm links: d = 450 − 35 − 10 − 25/2 = 392 mm This publication is available for purchase from www.concretecentre.com 100 Flexure in span: K = M Ed /bd 2 f ck = 184.4 × 10 6 /(300 × 392 2 × 30) = 0.133 Fig. 3.5 z/d = 0.864 Appendix A1 z = 0.864 × 392 = 338 mm A s = M Ed /f yd z = 184.4 × 10 6 /(434.8 × 338) = 1255 mm 2 Try 3 no. H25 B (1473 mm 2 ) (r = 1.25%) Table C5 Check spacing: Spacing of outer bars = 300 – 2 × 35 − 2 × 10 – 25 = 185 mm Assuming 10 mm diameter link, = spacing = 98 mm Steel stress under quasi-permanent loading: s s = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ULS loads) (1/d) = f yd × (A s,req /A s,prov ) × (g k + c 2 q k )/(g G g k + g Q q k ) (1/d) = (500/1.15) × (1255/1473) × [(30.2 + 0.3 × 11.5)/50.8] (1/1.03) = 434.8 × 0.91 × 0.66 × 0.97 = 237 MPa Cl. 7.3.3(2) As exposure is XC3, max. crack width w max = 0.3 mm Cl. 7.3.1(5) & NA = Maximum bar size = 16 mm or max. spacing = 200 mm = OK = Use 3 H25 B (1473 mm 2 ) Table 7.2N & NA Deflection: Check span-to-effective-depth ratio. Appendix B Basic span: effective depth ratio for r = 1.25%: Table 7.4N & NA l/d = 18 + [(1.25 – 0.5)/(1.5/0.5)] × (26 – 18) = 24.0 Max. span = 24.0 × 392 = 9408 mm = OK Flexure, support: M Ed = 193.8 kNm K = M Ed /bd 2 f ck where d = 450 − 35 − 10 − 25/2 = 392 mm K = 193.8 × 10 6 /(300 × 392 2 × 30) = 0.142 By inspection, K ≤ K' (0.142 × 0.168 ‡ ) = no compression reinforcement required. z = 0.85d = 0.85 × 392 = 333 mm A s = M Ed /f yd z = 193.8 × 10 6 /434.8 × 333 = 1338 mm 2 Try 3 no. H25 T (1473 mm 2 ) (r = 1.13%) Appendix A1 Table C5 ‡ K' is limited to 0.208. However, if, as is usual practice in the UK, x/d is limited to 0.45, z/d is as a consequence limited to 0.82 and K' to 0.168. This publication is available for purchase from www.concretecentre.com 4! Cont|nuous |eem on p|n supports 101 4.1.6 Shear a) Support B (critical) Shear at central support = 192.0 kN At d from face of support § V Ed = 192.0 − (0.300/2 + 0.392) × 50.8 = 164.50 kN Cl. 6.2.1(8) v Ed = V Ed /bd = 164.5 × 10 3 /(392 × 300) = 1.40 MPa Maximum shear capacity: Assuming f ck = 30 MPa and cot y = 2.5 # v Rd,max * = 3.64 MPa Table C7 v Rd,max > v Ed = OK Shear reinforcement: Assuming z = 0.9d Cl. 6.2.3(1) A sw /s ≥ V Ed /(0.9d × f ywd × cot y) ≥ 164.5 × 10 3 /(0.9 × 392 × (500/1.15) × 2.5) = 0.429 Cl. 6.2.3(3), Exp. (6.8) More accurately, A sw /s ≥ V Ed /(z × f ywd × cot y) ≥ 164.5 × 10 3 /(333 × 1087) = 0.454 Cl. 6.2.3(3), Exp. (6.8) Minimum shear links, A sw,min /s = 0.08b w f ck 0.5 /f yk = 0.08 × 300 × 30 0.5 /500 = 0.263. Not critical Cl. 9.2.2(5) Max. spacing = 0.75d = 0.75 × 392 = 294 mm Cl. 9.2.2(6) Use H8 @ 200 (A sw / s = 0.50) b) Support A (and C) Shear at end support = 137.2 kN At face of support, V Ed = 137.2 − (0.150 + 0.392) × 50.8 = 109.7 kN Cl. 6.2.1(8) By inspection, shear reinforcement required and cot y = 2.5 Fig. C1a) A sw /s ≥ V Ed /(z × f ywd × cot y) ≥ 109.7 × 10 3 /[353 × (500/1.15) × 2.5] = 0.285 Use H8 @ 200 (A sw / s = 0.50) throughout ‡ Appendix C5.3 § Where applied actions are predominantly uniformly distributed, shear may be checked at d from the face of support. See also Section 4.2.11. # The absolute maximum for v Rd,max (and therefore the maximum value of v Ed ) would be 5.28 MPa when cot y would equal 1.0 and the variable strut angle would be at a maximum of 45°. * For determination of V Rd,max see Section 4.2.10. ‡ As maximum spacing of links is 294 mm, changing spacing of links would appear to be of limited benefit. Cl. 6.2.1(8) This publication is available for purchase from www.concretecentre.com 102 4.1.7 Summary of design A B C 3H25B 3H25B 3H25 H8 @ 200 centres in 2 legs Figure 4.3 Continuous rectangular beam: Summary of design Commentary It is usually presumed that the detailer would take the design summarised above and detail the beam to normal best practice [8,9] . The design would go no further where standard detailing is all that is required. Where the element is non-standard (e.g. where there are point loads), it should be incumbent on the designer to give the detailer specific information about curtailment, laps, etc. as illustrated below. The detailer’s responsibilities, standards and timescales should be clearly defined but it would be usual for the detailer to draw and schedule not only the designed reinforcement but all the reinforcement required to provide a compliant and buildable solution. The work would usually include the checking the following aspects and providing appropriate detailing: Minimum areas t Curtailment lengths t Anchorages t Laps t U-bars t Rationalisation t Critical dimensions t Details and sections t The determination of minimum reinforcement areas and curtailment lengths, using the principles in Eurocode 2 is shown below. In practice these would be determined from published tables of data or by using reference texts [8,9] . Nonetheless, the designer should check the drawing for design intent and compliance with the standards. It is therefore necessary for the designer to understand and agree the principles of the detailing used. 4.1.8 Detailing checks a) Minimum areas A s,min = 0.26(f ctm /f yk )b t d ≥ 0.0013b t d where Cl. 9.2.1.1 This publication is available for purchase from www.concretecentre.com 4! Cont|nuous |eem on p|n supports 103 b t = width of tension zone f ctm = 0.30 × f ck 0.666 A s,min = 0.26 × 0.30 × 30 0.666 × 300 × 392/500 = 177 mm 2 Table 3.1 b) Curtailment of main bars How to: Detailing Bottom: curtail 75% main bars 0.08l from end support = 480 mm say 450 mm from A 70% main bars 0.30l – a l = 0.3 × 6000 − 1.125 × d = 1800 − 1.125 × 392 = 1359 mm say 1350 from A Top: curtail 40% main bars 0.15l + a l = 900 + 441 = 1341 mm say 1350 from B 65% main bars 0.30l + a l = 1800 + 441 = 2241 mm say 2250 from B At supports: 25% of A s to be anchored at supports Cl. 9.2.1.2.(1), 9.2.1.4(1) & NA 25% of 1225 mm 2 = 314 mm 2 Use min. 2 no. H16 (402 mm 2 ) at supports A, B and C Cl. 9.2.1.5(1) In accordance with SMDSC [9] detail MB1 lap U-bars tension lap with main steel = 780 mm (in C30/37 concrete, H12, ‘poor’ bond condition) How to: Detailing = say 800 mm c) Summary of reinforcement details 750 450 2H16B 2H16 U-bars 2H12T Links omitted for clarity 2H12T 2H12 2H12 2H12 2H12 2H25 2H25 1H25 1H25 3H25 2H16 2H12 2H12 2H16 2H25 2H25 2H25 1H25 1H25 2H16 U-bars 3H25B 3H25B B A C 600 600 800 800 1350 1250 1350 100 100 800 800 800 800 750 450 3H25 Figure 4.4 Continuous rectangular beam: reinforcement details Note Subsequent detailing checks may find issues with spacing rules especially if the 'cage and splice bar' method of detailing were to be used. 2H32s T&B would be a suitable alternative to 3H25s T&B. This publication is available for purchase from www.concretecentre.com 104 4.2 Heavily loaded L-beam g k1 = 46.0 q k1 = 63.3 C B A 9000 8000 2000 G k2 = 88.7 Q k2 = 138.7 Figure 4.5 Heavily loaded L-beam This edge beam supports heavy loads from storage loads. The variable point load is independent of the variable uniformly distributed load. The beam is supported on 350 mm square columns 4000 mm long. f ck = 30 MPa; f yk = 500 MPa. The underside surface is subject to an external environment and a 2-hour fire resistance requirement. The top surface is internal and subject to a 2-hour fire resistance requirement. Assume that any partitions are liable to be damaged by excessive deflections. 750 350 b eff Figure 4.6 Section through L-beam 4.2.1 Actions Permanent: UDL from slab and cladding g k = 46.0 kN/m Point load from storage area above = 88.7 kN Variable: From slab q k = 63.3 kN/m Point load from storage area above = 138.7 kN lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Heavily loaded L-beam This publication is available for purchase from www.concretecentre.com 105 4.2.2 Cover a) Nominal cover, c nom , underside and side of beam c nom = c min + Dc dev Exp. (4.1) where c min = max[c min,b ; c min,dur ] where c min,b = minimum cover due to bond Cl. 4.4.1.2(3) = diameter of bar. Assume 32 mm main bars and 10 mm links c min,dur = minimum cover due to environmental conditions. Assuming primarily XC3/XC4 exposure (moderate humidity or cyclic wet and dry); secondarily XF1 exposure (moderate water saturation without de-icing salt, vertical surfaces exposed to rain and freezing) and C30/37 concrete, c min,dur = 25 mm Table 4.1 BS 8500-1: Table A4 Dc dev = allowance in design for deviation. Assuming no measurement of cover Dc dev = 10 mm = c nom = 32 + 10 = 42 mm to main bars or = 25 + 10 = 35 mm to links Use c nom = 35 mm to links (giving c nom = 45 mm to main bars) Cl. 4.4.1.2(3) b) Fire Check adequacy of section for 2 hours fire resistance REI 120. EC2-1-2: 5.6.3 By inspection, web thickness OK. EC2-1-2: Table 5.6 Axis distance, a, required = 35 mm OK by inspection. EC2-1-2: Table 5.6 = Try 35 mm nominal cover bottom and sides to 10 mm link. Nominal cover, c nom , top: By inspection, c nom = c min + Dc dev where c min = max[c min,b ; c min,dur ] where Exp. (4.1) c min,b = minimum cover due to bond = diameter of bar. Assume 32 mm main bars and 10 mm links Cl. 4.4.1.2(3) c min,dur = minimum cover due to environmental conditions. Assuming primarily XC1 and C30/37 concrete, c min,dur = 15 mm Table 4.1 BS 8500-1: Table A4 4? leev||y |oeded l-|eem This publication is available for purchase from www.concretecentre.com 106 4? leev||y |oeded l-|eem Dc dev = allowance in design for deviation. Assuming no measurement of cover Dc dev = 10 mm Cl. 4.4.1.2(3) = c nom = 32 + 10 = 42 mm to main bars or = 15 + 10 = 25 mm to links Use c nom = 35 mm to links (giving c nom = 45 mm to main bars) 4.2.3 Idealisation, load combination and arrangement Load combination: As loads are from storage, Exp. (6.10a) is critical. Table 2.5; ECO: A1.2.2, NA & Exp. (6.10a) Idealisation: This element is treated as a continuous beam framing into columns 350 × 350 ‡ × 4000 mm long columns below. Cl. 5.3.1(3) Arrangement: Choose to use all-and-alternate-spans-loaded. Cl. 5.1.3(1) & NA: Table NA.1 (option b) 4.2.4 Analysis Analysis by computer (spreadsheet TCC 41 Continuous Beam (A+D).xls in RC spreadsheets V.3 [28] assuming frame action with 350 mm square columns 4 m long fixed at base. Beam inertia based on T-section, b eff wide) with 15% redistribution at central support, limited redistribution of span moment and consistent redistribution of shear. ECO: A1.2.2 & NA; Cl. 5.3.1 (6) Table 4.2 Elastic and redistributed moments, kNm Span number 1 2 Elastic M 1168 745 Redistributed M 1148 684 d 0.98 0.92 2000 1394 kNm 195 kNm 108 kNm – 684 kNm –1148 kNm 1500 1000 500 –500 –1000 –1500 0 A B C Figure 4.7 Redistributed envelope, kNm ‡ Note: 350 × 350 is a minimum for columns requiring a fire resistance of 120 minutes. EC2-1-2: Table 5.2a This publication is available for purchase from www.concretecentre.com 107 1000 646 kN 794 kN – 499 kN – 1098 kN 500 – 500 – 1000 – 1500 0 A B C 4.2.5 Flexural design, support A M Ed = 195 kNm in hogging M Ed,min = 1148 × 0.25 in hogging and in sagging Cl. 9.2.1.2(1), 9.2.1.4(1) & NA = 287 kNm K = M Ed /bd 2 f ck where b = b eff = b eff1 + b w + b eff2 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 1 where b 1 = distance between webs/2 l 0 = nominal: assume 0 § = b eff1 = 0 mm = b eff2 Fig. 5.2 =b = b w = 350 mm d = 750 − 35 – 10 – 32/2 = 689 mm assuming 10 mm link and H32 in support. f ck = 30 MPa K = 287 × 10 6 /(350 × 689 2 × 30) = 0.058 Restricting x/d to 0.45 Appendix A1 K' = 0.168 K ≤ K' = section under-reinforced and no compression reinforcement required. z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d = (689/2) (1 + 0.89) ≤ 0.95 × 689 = 652 ≤ 654 = z = 652 mm A s = M Ed /f yd z Appendix A1 § The distance l 0 is described as the distance between points of zero moment, ‘which may be obtained from Figure 5.2’. In this case l 0 = 0. (see Figure 4.11). Cl. 5.3.2.1(2) Fig. 5.2 Fig. 4.11 This publication is available for purchase from www.concretecentre.com 108 where f yd = 500/1.15 = 434.8 MPa = 287 × 10 6 /(434.8 × 652) = 1012 mm 2 Try 2 no. H32 U-bars (1608 mm 2 ) Check anchorage of H32 U-bars. Bars need to be anchored distance ‘A’ into column SMDSC: 6.4.2 U-bar A Figure 4.9 Distance A Assuming column uses 35 mm cover, 10 mm links and 32 mm bars: Distance A = 2 [350 − 2 (35 + 10) ] − 32/2 − 32/2 + 750 – [2 (35 + 10)] − 2 × 32/2 – (4 − π) (3.5 + 0.5) × 32 SMDSC [9] , BS 8666 [19] : = 488 + 628 – 110 = 1006 mm Table 2 Anchorage length, l bd = al b,rqd ≥ l b,min Cl. 8.4.4, Exp. (8.4) where a = conservatively 1.0 l b,rqd = (f/4) (s sd /f bd ) Exp. (8.3) where f = 32 s sd = design stress in the bar at the ULS = 434.8 × 1012/1608 = 274 MPa f |d = ultimate bond stress = 2.25 n 1 n 2 f ct,d Cl. 8.4.2 (2) This publication is available for purchase from www.concretecentre.com 109 where n 1 = 1.0 for good bond conditions n 2 = 1.0 for bar diameter ≤ 32 mm f ct,d = a ct f ctk / g C = 1.0 × 2.0/1.5 = 1.33 MPa f bd = 2.25 × 1.33 = 3.0 MPa Cl. 3.1.6 (2), Tables 3.1 & 2.1, & NA l b,rqd = (32/4) (274/3.0) = 731 mm ‡ l b,min = max[10f; 100 mm] = 250 mm = l bd = 731 mm i.e. < 1006 mm = OK Use 2 no. H32 U-bars 4.2.6 Flexural design, span AB a) Span AB – Flexure M Ed = 1148 kNm K = M Ed /bd 2 f ck where b = b eff = b eff1 + b w + b eff2 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 1 where b 1 = distance between webs/2 Assuming beams at 7000 mm cc = (7000 – 350)/2 = 3325 mm l 0 = 0.85 × l 1 = 0.85 × 9000 = 7650 mm § Fig. 5.2 b eff b w b w b 1 b 1 b 2 b 2 b b eff,1 b eff,2 Figure 4.10 Effective flange width b eff b Fig. 5.3 ‡ Anchorage lengths may be obtained from published tables. In this instance, a figure of 900 mm may be obtained from Table 13 of Section 10 of How to design concrete structures using Eurocode 2. § The distance l 0 is described as the distance between points of zero shear, which may be obtained from Figure 5.2’. From the analysis, l 0 could have been taken as 7200 mm. How to: Detailing [8] Cl. 5.3.2.1(2) Figure 5.2 4? leev||y |oeded l-|eem This publication is available for purchase from www.concretecentre.com 110 I 0 = 0.85 I 1 I 0 = 0.15 (I 1 +I 2 ) I 0 = 0.15 (I 2 +I 3 ) I 3 I 2 I 1 I 0 = 0.7 I 2 l 0 for calculation of flange width Fig. 5.2 b eff1 = 0.2 × 3325 + 0.1 × 7650 ≤ 0.2 × 7650 ≤ 3325 = 1430 ≤ 1530 ≤ 3325 = 1430 mm b w = 350 mm b eff2 = (0.2b 2 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 2 where b 2 = 0 mm b eff2 = 0 mm b = 1430 + 350 + 0 = 1780 mm d = effective depth = 750 − 35 – 10 – 32/2 = 689 mm assuming 10 mm link and H32 in span f ck = 30 MPa K = 1148 × 10 6 /(1780 × 689 2 × 30) = 0.045 Restricting x/d to 0.45, Appendix A1 K' = 0.168 K ≤ K' = section under-reinforced and no compression reinforcement required. z = lever arm = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d = (689/2) (1 + 0.917) ≤ 0.95 × 689 = 661 ≤ 654 = z = 654 mm Appendix A1 But z = d – 0.4x = by inspection, neutral axis is in flange and as x < 1.25 h f , design as rectangular section. A s = M Ed /f yd z where f yd = 500/1.15 = 434.8 MPa = 1148 × 10 6 /(434.8 × 654) = 4037 mm 2 Try 5 no. H32 B (4020 mm 2 ) (say OK) Appendix A1 Check spacing of bars. Spacing of bars = [350 – 2 × (35 + 10) – 32]/(5 – 1) = 57 Clear spacing = 57 – 32 mm = 25 mm between bars This publication is available for purchase from www.concretecentre.com 111 Minimum clear distance between bars = max[bar diameter; aggregate size + 5 mm] = max[32; 20 + 5] = 32 mm i.e. > 25 mm = 5 no. H32 B no good For 4 bars in one layer, distance between bars = 44 mm so Try 4 no. H32 B1 + 2 no. H32 B3 Cl. 8.2(2) & NA d 300 350 35 cover 32 bar 32 spacers 32 bar 10 link 35 cover 400 Figure 4.12 Span AB bottom reinforcement d = 750 − 35 – 10 – 32/2 – 0.333 × 2 × 32 = 668 mm K = 1148 × 10 6 /(1780 × 668 2 × 30) = 0.048 K ≤ K' = section under-reinforced and no compression reinforcement required. Appendix A1 z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d = (668/2) (1 + 0.911) ≤ 0.95 × 668 = 639 ≤ 635 = z = 635 mm Appendix A1 = by inspection, neutral axis in flange so design as rectangular section. A s = M Ed /f yd z = 1148 × 10 6 /(434.8 × 635) = 4158 mm 2 = 4 no. H32 B1 + 2 no. H32 B3 (4824 mm 2 ) OK Appendix A1 b) Span AB – Deflection Check end span-to-effective-depth ratio. Appendix B Allowable l/d = N × K × F1 × F2 × F3 where Appendix C7 N = Basic l/d: check whether r > r 0 and whether to use Exp. (7.16a) or (7.16b) Cl. 7.4.2(2), Exp. (7.16a), Exp. (7.16b) 4? leev||y |oeded l-|eem This publication is available for purchase from www.concretecentre.com 112 r = A s /A c ‡ = A s,req /[b w d + (b eff – b w )h f ] PD 6687 [6] = 4158/[350 × 668 + (1780 – 350) × 300] = 4158/662800 = 0.63% r 0 = f ck 0.5 /1000 = 30 0.5 /1000 = 0.55% r > r 0 = use Exp. (7.16b) N = 11 + 1.5f ck 0.5 r 0 /(r – r' ) + f ck 0.5 (r' /r 0 ) 0.5 /12 = 11 + 1.5 (30 0.5 × 0.055/(0.063 – 0) + 30 0.5 (0/0.55) 1.5 = 11 + 7.2 + 0 = 18.2 Exp. (7.16b) K = (end span) = 1.3 Table 7.4N & NA F1 = (b eff /b w = 1780/350 = 5.1) = 0.80 Cl. 7.4.2(2), Appendix C7 F2 = 7.0/l eff (span > 7.0 m) Cl. 7.4.2(2) where l eff = 9000 mm Cl. 5.3.2.2(1) F2 = 7.0/9.0 = 0.77 F3 = 310/s s ≤ 1.5 Cl. 7.4.2, Exp. (7.17), Table 7.4N & NA, Table NA.5 Note 5 where s s in simple situations = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ ULS loads) (1/d). However in this case separate analysis at SLS would be required to determine s s . Therefore as a simplification use the conservative assumption: Appendix B 310/s s = (500/f yk ) (A s,req /A s,prov ) = (500/500) × (4824/4158) = 1.16 Exp. (7.17) = Permissible l/d = 18.2 × 1.3 × 0.80 × 0.77 × 1.16 = 16.9 Actual l/d = 9000/668 = 13.5 Permissible more than actual = OK = 4 no. H32 B1 + 2 no. H32 B3 (4824 mm 2 ) OK 4.2.7 Flexural design, support B At centreline of support B, M = 1394 kNm From analysis, at face of support M EdBA = 1209 kNm M EdBC = 1315 kNm K = M Ed /b w d 2 f ck Cl. 5.3.2.2(3) ‡ 2.18 of PD 6687 [6] suggests that r in T sections should be based on the area of concrete above the centroid of the tension steel. This publication is available for purchase from www.concretecentre.com 113 where b w = 350 mm d = 750 − 35 – 12 – 32/2 = 687 mm assuming 10 mm link and H32 in support but allowing for H12 T in slab f ck = 30 MPa = K = 1315 × 10 6 /(350 × 687 2 × 30) = 0.265 for d = 0.85, K' = 0.168: to restrict x/d to 0.45, K' = 0.167 = Compression steel required Appendix A1 Table C4 z = (d/2) [1 + (1 − 3.53 K' ) 0.5 ] = (687/2) [1 + (1 − 3.53 × 0.167) 0.5 ] = (687/2) (1 + 0.64) < 0.95d = 563 mm A s2 = (K – K' )f ck bd 2 /f sc (d − d 2 ) where d 2 = 35 + 10 + 32/2 = 61 mm f sc = 700(x − d 2 )/x < f yd where x = 2.5 (d – z) = 2.5 (687 – 563) = 310 mm f sc = 700 × (310 − 61)/310 < 500/1.15 = 562 MPa but limited to ≤ 434.8 MPa =A s2 = (0.265 – 0.167) × 30 × 350 × 687 2 /[434.8(687 − 61) ] = 1784 mm 2 Try 4 no. H25 B (1964 mm 2 ) Fig. 3.5, Appendix A1, How to: Beams A s = M'/f yd z + A s2 f sc /f yd = K' f ck bd 2 /(f yd z) + A s2 f sc /f yd = 0.167 × 30 × 350 × 687 2 /(434.8 × 563) + 1570 × 434.8/434.8 = 3380 + 1784 = 5164 mm 2 Try 4 no. H32 T + 4 no. H25 T (5180 mm 2 ) Appendix A1 This reinforcement should be spread over b eff . b eff = b eff1 + b w + b eff2 Cl. 9.2.1.2(2), Fig. 9.1 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 1 where b 1 = distance between webs/2. Assuming beams at 7000 mm cc = (7000 – 350)/2 = 3325 mm l 0 = 0.15 × (l 1 + l 2 ) Fig. 5.2 = 0.15 × (9000 + 8000) = 2550 mm 4? leev||y |oeded l-|eem This publication is available for purchase from www.concretecentre.com 114 =b eff1 = 0.2 × 3325 + 0.1 × 2550 ≤ 0.2 × 2550 ≤ 3325 = 920 ≤ 510 ≤ 3325 = 510 mm b w = 350 mm b eff2 = (0.2b 2 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 2 where b 2 = 0 mm b eff2 = 0 mm = b eff = 510 + 350 + 0 = 860 mm Use 4 no. H32 T + 4 no. H25 T (5180 mm 2 ) @ approx 100 mm cc b w = 350 mm 4H25 4H25 4H32 b eff = 860 mm Figure 4.13 Support B reinforcement pp 4.2.8 Flexural design, span BC a) Span BC – Flexure M Ed = 684 kNm K = M Ed /bd 2 f ck where b = b eff = b eff1 + b w + b eff2 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2l 0 ≤ b 1 where b 1 = distance between webs/2. Assuming beams at 7000 mm cc = (7000 – 350)/2 = 3325 mm Fig. 5.2 l 0 = 0.85 × l 1 = 0.85 × 8000 = 6800 mm b eff1 = 0.2 × 3325 + 0.1 × 6800 ≤ 0.2 × 6800 ≤ 3325 = 1345 ≤ 1360 ≤ 3325 = 1360 mm This publication is available for purchase from www.concretecentre.com 115 b w = 350 mm b eff2 = (0.2b 2 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 2 where b 2 = 0 mm b eff2 = 0 mm = b = 1360 + 350 + 0 = 1710 mm d = 750 − 35 – 10 – 32/2 = 689 mm assuming 10 mm link and H32 in span. f ck = 30 MPa = K = 684 × 10 6 / (1710 × 689 2 × 30) = 0.028 By inspection, K ≤ K' = section under-reinforced and no compression reinforcement required. Appendix A1 z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d = (689/2) (1 + 0.95) ≤ 0.95 × 689 = 672 > 655 = z = 655 mm Appendix A1 By inspection, x < 1.25 h f ; design as rectangular section A s = M Ed /f yd z = 684 × 10 6 /(434.8 × 655) = 2402 mm 2 Try 2 no. H32 B + 2 no. H25 B (2590 mm 2 ) Appendix A1 b) Span BC – Deflection By inspection, compared with span AB OK 4.2.9 Flexural design, support C By inspection, use 2 no. H25 U-bars as support A. Use 2 no. H25 U-bars 4.2.10 Design for beam shear, support A At d from face of support V Ed = 646 − (350/2 + 0.689) × (1.35 × 46.0 + 1.5 × 63.3) Cl. 6.2.1(8) ECO: A1.2.2, NA & Exp. (6.10a) = 646 – 0.864 × 157.1 = 510.3 kN Check maximum shear resistance. V Rd, max = a cw b w zvf cd /(cot y + tan y) Exp. (6.9) & NA where a cw = 1.0 Cl. 6.2.3 & NA b w = 350 mm as before z = 0.9d Cl. 6.2.3(1) 4? leev||y |oeded l-|eem This publication is available for purchase from www.concretecentre.com 116 v = 0.6 (1 − f ck /250) = 0.6 (1 − 30/250) = 0.528 Cl. 6.2.3(3) Note 1, Exp. (6.6N) & NA f cd = 30/1.5 = 20.0 MPa Cl. 2.4.2.4(1) & NA y = angle of inclination of strut. = 0.5 sin −1 {v Ed,z /[0.20 f ck (1 – f ck /250) ] } ≥ cot −1 2.5 Exp. (6.9), Appendix A2 where v Ed,z = V Ed /b z = V Ed /(b × 0.9d) = 510.3 × 10 3 /(350 × 0.9 × 689) = 2.35 MPa y = 0.5 sin −1 {2.35/[0.20 × 30 (1 – 30/250) ] } ≥ cot −1 2.5 = 0.5 sin −1 (0.445) ≥ cot −1 2.5 = 0.5 × 26.4° ≥ 21.8° = 21.8° = V Rd,max = 1.0 × 350 × 0.90 × 689 × 0.528 × 20.0/(2.5 + 0.4) = 790 kN = OK Shear reinforcement: Shear links: shear resistance with links V Rd,s = (A sw /s) z f ywd cot y = A sw /s ≥ V Ed /z f ywd cot y Exp. (6.8) where A sw /s = area of legs of links/link spacing z = 0.9d as before f ywd = 500/1.15 = 434.8 Cl. 2.4.2.4(1) & NA cot y = 2.5 as before A sw /s ≥ 510.3 × 10 3 /(0.9 × 689 × 434.8 × 2.5) = 0.76 Minimum A sw /s = r w,min b w sin a Cl. 9.2.2(5), Exp. (9.4) where r w,min = 0.08 × f ck 0.5 /f yk = 0.08 × 30 0.5 /500 Exp. (9.5N) & NA = 0.00088 b w = 350 mm as before a = angle between shear reinforcement and the longitudinal axis. For vertical reinforcement sin a = 1.0 = Minimum A sw /s = 0.00088 × 350 × 1 = 0.03 But, maximum spacing of links longitudinally = 0.75d = 516 mm = Try H10 @ 200 cc in 2 legs (A sw /s = 0.78) Cl. 9.2.2(6) 4.2.11 Design for high beam shear, support B As uniformly distributed load predominates consider at d from face of support. Cl. 6.2.1(8) This publication is available for purchase from www.concretecentre.com 117 V Ed = 1098 − (350/2 + 0.689) × (1.35 × 46.0 + 1.5 × 63.3) = 1098 – 0.864 × 157.1 = 962.3 kN By inspection, shear reinforcement required and cot y < 2.5. Check V Rd, max (to determine y) Check maximum shear resistance. As before, V Rd, max = a cw b w zvf cd /(cot y + tan y). Exp. (6.9) & NA where a cw , b w , z, v and f cd as before y = 0.5 sin −1 {v Ed,z /[0.20 f ck (1 – f ck /250) ] } ≥ cot −1 2.5 Exp. (6.9) where v Ed,z = V Ed /bz = V Ed /(b0.9d) = 962.3 × 10 3 /(350 × 0.9 × 687) = 4.45 MPa Cl. 6.2.3(1) y = 0.5 sin −1 {4.45/[0.20 × 30 (1 – 30/250) ] } ≥ cot −1 2.5 = 0.5 sin −1 (0.843) ≥ cot −1 2.5 = 0.5 × 57.5° ≥ 21.8° = 28.7° Exp. (6.9) cot y = 1.824 i.e. > 1.0 =OK Cl. 6.2.3(2) & NA tan y = 0.548 = V Rd,max = 1.0 × 350 × 0.90 × 687 × 0.528 × 20.0/(1.824 + 0.548) = 963.4 kN OK (i.e. V Rd,max ≈ V Ed ) Shear reinforcement: Shear links: shear resistance with links V Rd,s = (A sw /s)zf ywd cot y = A sw /s ≥ V Ed /zf ywd cot y Exp. (6.8) A sw /s ≥ 962.3 × 10 3 /(0.9 × 687 × 434.8 × 1.824) = 1.96 = Use H10 @ 150 cc in 4 legs (A sw /s = 2.09) 4.2.12 Design for beam shear (using design chart), support B C At d from face of support, V Ed = 794 – 0.864 × 157.1 = 658.3 kN v Ed,z = V Ed /bz = V Ed /(b0.9d) = 658.3 × 10 3 / (350 × 0.9 × 687) = 3.04 MPa Cl. 6.2.1(8) From chart A sw /s reqd /m width = 2.75 A sw /s reqd = 2.75 × 0.35 = 0.96 = Use H10 in 2 legs @ 150 mm cc (A sw /s = 1.05) Fig. C1b) 4? leev||y |oeded l-|eem This publication is available for purchase from www.concretecentre.com 118 4.2.13 Check shear capacity for general case In mid span use H10 in 2 legs @ 300 mm cc (A sw /s = 0.52) z A sw /s reqd /m width = 1.48 and an allowable v Ed,z = 1.60 MPa z 1.60 × 350 × 0.90 × 687 = V Ed = 346 kN From analysis, V Ed = 346.2 kN occurs at: (646 − 346)/157.1 = 1900 mm from A, (1098 – 346 − 1.25 × 88.7 – 1.5 × 138.7)/157.1 = 2755 mm from B A , (794 − 346)/157.1 = 2850 mm from B C and (499 − 346)/157.1 = 970 mm from C Fig. C1b) 4.2.14 Summary of design 2H32 U-bars A 4H32 + 4H25T 2H25 U-bars H10 links in 2 legs 4H32 B1 + 2H32 B3 @ 300 @ 300 @ 150 @ 150 @ 150 @ 150 H10 links in 2 legs @ 150cc 1 1 4H25B 2H32 + 2H25B 1050 2850 2850 1950 B C Figure 4.14 Summary of L-beam design 350 510 4H25 4H32 H10 in 4 legs @ 150 Figure 4.15 L-beam section 1–1 This publication is available for purchase from www.concretecentre.com 119 q k = 45.8 kN/m g k = 47.8 kN/m A B C D E 7500 7500 7500 7500 Figure 4.16 Continuous wide T-beam This central spine beam supports the ribbed slab in Example 3.3. The 300 mm deep ribbed slab is required for an office to support a variable action of 5 kN/m 2 . The beam is the same depth as the slab and is supported on 400 mm square columns, see Figure 4.17. f ck = 35 MPa; f yk = 500 MPa. A 1-hour fire resistance is required in an internal environment. Assume that partitions are liable to be damaged by excessive deflections. C L 100 200 200 800 1000 Figure 4.17 Section through T-beam 4.3 Continuous wide T-beam 4.3.1 Actions Permanent, UDL ‡ : From analysis of slab, g k = 47.8 kN/m Variable: From analysis of slab, q k = 45.8 kN/m ‡ The actions may also have been estimated assuming an elastic reaction factor of 1.1 for the slab viz: kN/m Permanent: UDL Loads from ribbed slab (7.50 + 9.0)/2 × 4.30 × 1.1 = 39.0 Self-weight/patch load extra over solid 2.0 × 4.17 = 8.3 47.3 Variable: Imposed (7.50 + 9.0)/2 × 5.00 × 1.1 = 45.4 lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Continuous wide T-beam 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 120 4.3.2 Cover Nominal cover, c nom : c nom = c min + Dc dev Exp. (4.1) where c min = max[c min,b ; c min,dur ] where c min,b = minimum cover due to bond Cl. 4.4.1.2(3) = diameter of bar. Assume 25 mm main bars and 8 mm links c min,dur = minimum cover due to environmental conditions. Assuming XC1 and C30/37 concrete, c min,dur = 15 mm Table 4.1 BS 8500-1; Table A4 Dc dev = allowance in design for deviation. Assuming no measurement of cover Dc dev = 10 mm Cl. 4.4.1.2(3) = c nom = 15 + 10 = 25 mm to links or = 25 + 10 = 35 mm to main bars Use 10 mm diameter links to give c nom = 35 mm to main bars and 25 mm to links (as per ribbed slab design). Fire: Check adequacy of section for REI 60. EC2-1-2: 5.6.3 EC2-1-2: Table 5.6 Axis distance required: Minimum width b min = 120 mm with a = 25 mm or b min = 200 mm with a = 12 mm EC2-1-2: Table 5.6 = at 2000 mm wide (min.) a < 12 mm By inspection, not critical. Use 25 mm nominal cover to links 4.3.3 Idealisation, load combination and arrangement Load combination: By inspection, Exp. (6.10b) is critical. 47.8 × 1.25 + 45.8 × 1.5 = 128.5 kN/m ‡ Fig. 2.5 EC0: Exp. (6.10b) Idealisation: This element is treated as a beam on pinned supports. The beam will be provided with links to carry shear and to accommodate the requirements of Cl. 9.2.5 – indirect support of the ribbed slab described in Section 3.3.8. Arrangement: Choose to use all-and-alternate-spans-loaded. Cl. 5.1.3(1) & NA: Table NA.1 (option b) ‡ cf. 126.7 kN/m from analysis of slab (63.2 kN/m + 63.5 kN/m). See Figure 3.12. This publication is available for purchase from www.concretecentre.com 121 4.3.4 Analysis Analysis by computer, assuming simple supports and including 15% redistribution at supports (with in this instance consequent redistribution in span moments). EC0: A1.2.2 & NA; Cl. 5.3.1 (6) 5.3.1(6) Table 4.3 Elastic and redistributed moments, kNm Span number 1 2 3 4 Elastic M 641.7 433.0 433.0 641.7 Redistributed M 606.4 393.2 393.2 606.4 d 0.945 0.908 0.908 0.945 800 657.4 kNm 657.4 kNm 516.0 kNm – 606.4 kNm – 606.4 kNm – 393.2 kNm – 393.2 kNm 600 400 200 – 200 – 400 – 800 – 600 0 A B C D E Figure 4.18 Redistributed envelope, kNm 800 394.6 kN 569.1 kN 462.6 kN 517.9 kN – 517.9 kN – 569.1 kN – 395.6 kN – 462.6 kN 600 400 200 – 200 – 400 – 800 – 600 0 A B C D E Figure 4.19 Redistributed shears, kN 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 122 4.3.5 Flexural design, span AB a) Span AB (and DE) – Flexure M Ed = 606.4 kNm K = M Ed /bd 2 f ck where b = b eff = b eff1 + b w + b eff2 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2l 0 ≤ b 1 where b 1 = distance between webs/2 Referring to Figures 3.8 and 3.9 = (7500 – 1000 − 550)/2 = 2975 mm l 0 = 0.85 × l 1 = 0.85 × 7500 = 6375 mm Fig. 5.2 b eff1 = 0.2 × 2975 + 0.1 × 6375 ≤ 0.2 × 6375 ≤ 2975 = 1232 ≤ 1275 ≤ 2975 = 1232 mm b w = 2000 mm b eff2 = (0.2b 2 + 0.1l 0 ) ≤ 0.2 l 0 ≤ b 2 where b 2 = distance between webs/2. Referring to Figures 3.8 and 3.9 = (9000 – 1000 − 550)/2 = 3725 mm l 0 = 6375 mm as before b eff2 = 0.2 × 3725 + 0.1 × 6375 ≤ 0.2 × 6375 ≤ 3725 = 1382 ≤ 1275 ≤ 3725 = 1275 mm b = 1232 + 2000 + 1275 = 4507 mm d = 300 − 25 – 10 – 25/2 = 252 mm assuming 10 mm link and H25 in span. f ck = 35 MPa K = 606.4 × 10 6 /(4507 × 252 2 × 35) = 0.061 K' = 0.207 Appendix A1 or restricting x/d to 0.45 K' = 0.168 K ≤ K' = section under-reinforced and no compression reinforcement required. z = (d/2) [1 + (1 − 3.53K) 0.5 ] ≤ 0.95d Appendix A1 = (252/2) (1 + 0.886) ≤ 0.95 × 252 = 238 ≤ 239 = z =238 mm This publication is available for purchase from www.concretecentre.com 123 But z = d – 0.4 x Appendix A1 = x = 2.5(d – z) = 2.5( 252 − 236) = 32 mm = neutral axis in flange. A s x < 1.25h f design as rectangular section. A s = M Ed /f yd z where f yd = 500/1.15 = 434.8 MPa = 606.4 × 10 6 /(434.8 × 239) = 5835 mm 2 Try 12 no. H25 B (5892 mm 2 ) b) Span AB – Deflection Check span-to-effective-depth ratio. Appendix B Allowable l/d = N × K × F1 × F2 × F3 Appendix C7 where N = Basic l/d: check whether r > r 0 and whether to use Exp. (7.16a) or (7.16b) Cl. 7.4.2(2), Exp. (7.16a), Exp. (7.16b) r = A s /A c ‡ = A s,req /[b w d + (b eff – b w )h f ] PD 6687 [6] = 5835/[2000 × 252 + (4507 – 2000) × 100] = 5835/754700 = 0.77% r 0 = f ck 0.5 /1000 = 30 0.5 /1000 = 0.59% r > r 0 = use Exp. (7.16b) N = 11 + 1.5f ck 0.5 r 0 /(r – r') + f ck 0.5 (r'/r 0 ) 0.5 /12 = 11 + 1.5 (35 0.5 × 0.059/(0.077 – 0) + 35 0.5 (0/0.59) 1.5 = 11 + 6.8 + 0 = 17.8 Exp. (7.16b) K = (end span) = 1.3 Table 7.4N & NA F1 = (b eff /b w = 4057/2000 = 2.03) = 0.90 Cl. 7.4.2(2), Appendix C7 F2 = 7.0/l eff (span > 7.0 m) where l eff = 7100 + 2 × 300/2 = 7400 mm F2 = 7.0/7.4 = 0.95 Cl. 7.4.2(2), 5.3.2.2(1) F3 = 310/s s ≤1.5 Cl. 7.4.2, Exp. (7.17), Table 7.4N & NA, Table NA.5 Note 5 where § s s = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ULS loads) (1/d) = 434.8 × (5835/5892) [(47.8 + 0.3 × 45.8)/(1.25 × 47.8 + 1.5 × 45.8)] × (1/0.945) = 434.8 × 0.99 × 0.48 × 1.06 = 219 MPa ‡ 2.18 of PD 6687 [6] suggests that r in T sections should be based on the area of concrete above the centroid of the tension steel. § See Appendix B1.5 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 124 F3 = 310/s s = 310/219 = 1.41 = Permissible l/d = 17.8 × 1.3 × 0.90 × 0.95 × 1.41 = 27.9 Actual l/d = 7500/252 = 29.8 = no good Try 13 no. H25 B (6383 mm 2 ) F3 = 310/s s = 310/219 × 13/12 = 1.53 ‡ = say 1.50 = Permissible l/d = 17.8 × 1.3 × 0.90 × 0.95 × 1.50 = 29.7 Actual l eff /d = 7400/252 = 29.4 Say OK Use 13 no. H25 B (6383 mm 2 ) 4.3.6 Flexural design, support B (and D) At centreline of support: M = 657.4 kNm At face of support: M Ed = 657.4 – 0.2 × 517.9 + 0.202 × 128.5/2 = 657.4 – 101.0 = 556.4 kNm K = M Ed /b w d 2 f ck where b w = 2000 mm d = 300 − 25 cover − 12 fabric − 8 link – 16 bar − 25/2 bar = 226 mm Cl. 5.3.2.2(3) 25 cover 8 link 8 link 20 bar 25 cover 12 fabric 16 bar 25 cover 8 link 25 bar 16 bar Figure 4.20 Section at rib-beam interface K = 556.4 × 10 6 /(2000 × 226 2 × 35) = 0.156 By inspection, K < K' K' = 0.167 maximum (or for d = 0.85, K' = 0.168) = No compression steel required. Appendix A1 Table C.4 ‡ Both A s,prov /A s,req and any adjustment to N obtained from Exp. (7.16a) or Exp. (7.16b) is restricted to 1.5 by Note 5 to Table NA.5 in the UK NA. NA, Table NA.5 This publication is available for purchase from www.concretecentre.com 125 z = (226/2)[1 + (1 − 3.53 K' ) 0.5 ] = (226/2)[1 + (1 − 3.53 × 0.156) 0.5 ] = (226/2) (1 + 0.67) < 0.95d = 189 mm A s = M Ed /f yd z = 556.4 × 10 6 /(434.8 × 189) = 6770 mm 2 Try 14 no. H25 T (6874 mm 2 ) To be spread over b eff Cl. 9.2.1.2(2), Fig. 9.1 b eff = b eff1 + b w + b eff2 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2l 0 ≤ b 1 where b 1 referring to Figure 3.9 = (7500 – 1000 – 550)/2 = 2975 mm l 0 = 0.15 × (l 1 + l 2 ) = 0.15 × (7500 + 7500) = 2250 mm b eff1 = 0.2 × 2975 + 0.1 × 2250 ≤ 0.2 × 2250 ≤ 2975 = 820 ≤ 450 ≤ 2975 Fig. 5.2 = 450 mm b w = 2000 mm b eff2 = 450 mm as before = b eff = 450 + 2000 + 450 = 2900 mm Check cracking: Spacing = 2900 – 2 × (25 – 10 – 25/2)/(14 − 1) = 216 mm s s = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ULS loads) (1/d) = 434.8 × (6770/6874) [ (47.8 + 0.3 × 45.8)/ (1.25 × 47.8 + 1.5 × 45.8) × (1/0.85) = 434.8 × 0.98 × 0.48 × 1.18 = 241 MPa Cl. 7.3.3 As loading is the cause of cracking satisfy either Table 7.2N or Table 7.3N Cl. 7.3.3(2) & Note For w k = 0.4 and s s = 240 MPa max. spacing = 250 mm = OK Table 7.3N 4.3.7 Flexural design, span BC (and CD similar) a) Flexure M Ed = 393.2 kNm K = M Ed /bd 2 f ck where b = b eff = b eff1 + b w + b eff2 Cl. 5.3.2.1, Fig. 5.3 where b eff1 = (0.2b 1 + 0.1l 0 ) ≤ 0.2l 0 ≤ b 1 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 126 where b 1 referring to Figure 3.9 = (7500 – 1000 – 550)/2 = 2975 mm l 0 = 0.70 × l 2 = 0.7 × 7500 = 5250 mm b eff1 = 0.2 × 2975 + 0.1 × 5250 ≤ 0.2 × 5250 ≤ 2975 Fig. 5.2 = 1120 ≤ 1050 ≤ 2975 = 1050 mm b w = 2000 mm b eff2 = (0.2b 2 + 0.1l 0 ) ≤ 0.2l 0 ≤ b 2 where b 2 = distance between webs/2 Referring to Figures 3.8 and 3.9 = (9000 – 1000 – 550)/2 = 3725 mm l 0 = 5250 mm as before b eff2 = 0.2 × 3725 + 0.1 × 5250 ≤ 0.2 × 5250 ≤ 3725 = 1270 ≤ 1050 ≤ 3725 = 1270 mm b = 1050 + 2000 + 1270 = 4320 mm d = 252 mm as before assuming 10 mm link and H25 in span f ck = 30 K = 393.2 × 10 6 /(4320 × 252 2 × 35) = 0.041 By inspection, K ≤ K' = section under-reinforced and no compression reinforcement required. Appendix A1 z = (d/2) [1 + (1 – 3.53K)0.5] ≤ 0.95d = (252/2) (1 + 0.924) ≤ 0.95 × 252 = 242 > 239 = z = 239 mm Appendix A1 By inspection, x < 1.25 h f = design as rectangular section Appendix A1 A s = M Ed /f yd z = 393.2 × 10 6 /(434.8 × 239) = 3783 mm 2 Try 8 no. H25 B (3928 mm 2 ) b) Deflection By inspection, compared to span AB OK But for the purposes of illustration: Check span-to-effective-depth ratio. Appendix B Allowable l/d = N × K × F1 × F2 × F3 Appendix C7 where N = Basic l/d: check whether to use Exp. (7.16a) or (7.16b) Cl. 7.4.2(2) This publication is available for purchase from www.concretecentre.com 127 r 0 = 0.59% (for f ck = 35) r = A s /A c ‡ = A s,req /[b w d + (b eff – b w )h f ] where b w = 2000 mm r = 3783/(2000 × 252 + (4320 – 2000) × 100) = 3783/736000 = 0.51% r < r 0 = use Exp. (7.16a) N = 11 + 1.5f ck 0.5 r 0 /r + 3.2f ck 0.5 (r 0 /r – 1)1.5 = 11 + 1.5 × 35 0.5 × 0.059/0.051 + 3.2 × 35 0.5 (0.059/0.051 – 1) 1.5 = 11 + 10.2 + 23.5 = 17.8 = 44.7 Exp. (7.16a) K = (internal span) = 1.5 Table 7.4N & NA F1 = (b eff /b w = 4320/2000 = 2.16) = 0.88 Cl. 7.4.2(2), Appendix C7 F2 = 7.0/l eff = 7.0/7.4 = (span > 7.0 m) = 0.95 Cl. 7.4.2(2) F3 = 310/s s ≤1.5 Cl. 7.4.2, Exp. (7.17) Table 7.4N, & NA, Table NA.5 Note 5 where § s s = (f yk /g S ) (A s,req /A s,prov ) (SLS loads/ULS loads) (1/d) = 434.8 × (3783/3828) [(47.8 + 0.3 × 45.8)/(1.25 × 47.8 + 1.5 × 45.8)] × (1/0.908) = 434.8 × 0.99 × 0.48 × 1.10 = 227 MPa F3 = 310/ s s = 310/227 = 1.37 = Permissible l/d = 44.7 × 1.37 × 0.88 × 0.95 × 1.37 = 70.1 Actual l/d = 7500/252 = 29.8 = OK Use 8 no. H25 B (3928 mm 2 ) # c) Hogging Assuming curtailment of top reinforcement at 0.30l + a l , How to: Detailing From analysis M Ed at 0.3l from BC (& DC) = 216.9 kNm at 0.3l from CB (& CD) = 185.6 kNm K = 216.9 × 10 6 /(2000 × 226 2 × 35) = 0.061 By inspection, K < K' ‡ 2.18 of PD 6687 [6] suggests that r in T sections should be based on the area of concrete above the centroid of the tension steel. § See Appendix B1.5 # 12 no. H20 B (3768 mm 2 ) used to suit final arrangement of links. 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 128 z = (226/2)[1 + (1 − 3.53 K') 0.5 ] = (226/2)[1 + (1 − 3.53 × 0.061) 0.5 ] = (226/2) (1 + 0.89) < 0.95d = 214 mm < 215 mm A s = M Ed /f yd z = 216.9 × 10 6 /(434.8 × 214) = 2948 mm 2 Use 12 no. H20 T (3748 mm 2 ) (to suit links and bottom steel) Top steel at supports may be curtailed down to 12 no. H20 T at 0.3l + a l = 0.3 × 7500 + 1.25 × 214 = 2518 say 2600 mm from centreline of support. Cl. 9.2.1.3(2) 4.3.8 Flexural design, support C At centreline of support, M = 516.0 kNm At face of support, Cl. 5.3.2.2(3) M Ed = 516.0 – 0.2 × 462.6 + 0.20 2 × 128.5/2 = 516.0 – 90.0 = 426.0 kNm K = M Ed /b w d 2 f ck where b w = 2000 mm d = 226 mm as before K = 426.0 × 10 6 /(2000 × 226 2 × 35) = 0.119 By inspection, K < K' z = (226/2) [1 + (1 − 3.53K) 0.5 ] = (226/2) [1 + (1 − 3.53 × 0.119) 0.5 ] = (226/2) (1 + 0.76) < 0.95d = 199 mm A s = M Ed /f yd z = 426.0 × 10 6 /(434.8 × 199) = 4923 mm 2 Try 10 no. H25 T (4910 mm 2 ) ‡ 4.3.9 Design for beam shear a) Support A (and E) At d from face of support, V Ed = 394.6 − (0.400/2 + 0·252) × 128.5 = 336.5 kN Cl. 6.2.1(8) Maximum shear resistance: By inspection, V Rd,max OK and cot y = 2.5 ‡ 12 no. H25 used to suit final arrangement of links. This publication is available for purchase from www.concretecentre.com 129 However, for the purpose of illustration: check shear capacity, V Rd,max = a cw b w zvf cd / (cot y + tan y) where Exp. (6.9) & NA a cw = 1.0 b w = 2000 mm as before z = 0.9d v = 0.6 [1 − f ck /250] = 0.516 Cl. 6.2.3(1) f cd = 35/1.5 = 23.3 MPa y = angle of inclination of strut. By inspection, cot −1 y << 21.8. But cot y restricted to 2.5 and = tan y = 0.4. Cl. 6.2.3(2) & NA V Rd,max = 1.0 × 2000 × 0.90 × 252 × 0.516 × 23.3/(2.5 + 0·4) = 2089.5 kN = OK Shear links: shear resistance with links \ Rd,s = (A sw /s) z f ywd cot y ≥ V Ed Exp. (6.8) = for V Ed ≤ V Rd,s A sw /s ≥ V Ed /z f ywd cot y where A sw /s = area of legs of links/link spacing z = 0.9d as before f ywd = 500/1.15 = 434.8 cot y = 2.5 as before A sw /s ≥ 336.5 × 10 3 /(0.9 × 252 × 434.8 × 2.5) = 1.36 Minimum A sw /s = r w,min b w sin a Cl. 9.2.2(5), Exp. (9.4) where r w,min = 0.08 × f ck 0.5 /f yk = 0.08 × 35 0.5 /500 = 0.00095 Exp. (9.5N) & NA b w = 2000 mm as before a = angle between shear reinforcement and the longitudinal axis. For vertical reinforcement sin a = 1.0 Minimum A sw /s = 0.00095 × 2000 × 1 = 1.90 But, maximum spacing of links longitudinally = 0.75d = 183 mm Cl. 9.2.2(6) Maximum spacing of links laterally = 0.75d ≤ 600 mm = 183 mm H10s required to maintain 35 mm cover to H25 = Use H10 @ 175 cc both ways i.e. H10 in 12 § legs @ 175 mm cc (A sw /s = 5.38) Cl. 9.2.2(8) § (2000 mm – 2 × 25 mm cover − 10 mm diameter)/175 = 11 spaces, = 12 legs. 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 130 b) Support B (and C and D) By inspection, the requirement for minimum reinforcement and, in this instance, for H10 legs of links will outweigh design requirements. Nonetheless check capacity of A sw /s = 5.38 V Rd,s = (A sw /s) z f ywd cot y Exp. (6.8) = 5.38 × 0.9 × 252 × 434.8 × 2.5 = 1326.3 kN Maximum shear at support = 517.9 kN i.e. capacity of minimum links not exceeded. By inspection, the requirement for indirect support of the ribs of the slab using 87 mm 2 /rib within 150 mm of centreline of ribs (at 900 mm centres) and within 50 mm of rib/solid interface is adequately catered for and will not unduly effect the shear capacity of the beam. Use 150 mm centres to tie in with 900 mm centres of ribs = Use H10 in 12 legs @ 150 mm cc (A sw /s = 6.28) throughout beam Cl. 9.2.5, Section 3.4.8 4.3.10 Check for punching shear, column B As the beam is wide and shallow it should be checked for punching shear. At B, applied shear force, V Ed = 569.1 + 517.9 = 1087.0 kN. Check at perimeter of 400 × 400 mm column: v Ed = bV Ed /u i d < v Rd,max Cl. 6.4.3(2), 6.4.5(3) where b = factor dealing with eccentricity; recommended value 1.15 V Ed = applied shear force Fig. 6.21N & NA u i = control perimeter under consideration. For punching shear adjacent to interior columns u 0 = 2(c x + c y ) = 1600 mm Cl. 6.4.5(3) d = mean d = (245 + 226)/2 = 235 mm Exp. (6.32) v Ed = 1.15 × 1087.0 × 10 3 /1600 × 235 = 3.32 MPa v Rd,max = 0.5vf cd Cl. 6.4.5(3) Note where v = 0.6(1 − f ck /250) = 0.516 Exp. (6.6) & NA f cd = a cc lf ck /g C = 1.0 × 1.0 × 35/1.5 = 23.3 v Rd,max = 0.5 × 0.516 × 23.3 = 6.02 MPa = OK Table C7 ‡ Check shear stress at basic perimeter u 1 (2.0d from face of column): v Ed = bV Ed /u 1 d < v Rd,c where Cl. 6.4.2 b, V Ed and d as before Fig. 6.13 ‡ In this case, at the perimeter of the column, it is assumed that the strut angle is 45°, i.e. that cot y = 1.0. In other cases, where cot y < 1.0, v Rd,max is available from Table C7. This publication is available for purchase from www.concretecentre.com 131 u 1 = control perimeter under consideration. For punching shear at 2d from interior columns = 2(c x + c y ) + 2π × 2d = 1600 + 2π × 2 × 235 = 4553 mm v Ed = 1.15 × 1087.0 × 10 3 /4553 × 235 = 1.17 MPa v Rd,c = 0.18/ g C × k × (100 r l f ck ) 0.333 Exp. (6.47) & NA where g C = 1.5 k = 1 + (200/d) 0.5 ≤ 2 = 1 +(200/235) 0.5 = 1.92 r l = (r ly, r lz ) 0.5 Cl. 6.4.4.1(1) where r ly, r lz = Reinforcement ratio of bonded steel in the y and z direction in a width of the column plus 3d each side of column. = 6874/(2000 × 226) = 0.0152 r lz = 741/(900 × 245) = 0.0036 r l = (0.0152 × 0.0036) 0.5 = 0.0074 f ck = 35 v Rd,c = 0.18/1.5 × 1.92 × (100 × 0.0074 × 35) 0.333 = 0.68 MPa § Table C6 # = punching shear reinforcement required Shear reinforcement (assuming rectangular arrangement of links): At the basic control perimeter, u 1 , 2d from the column: A sw ≥ (v Ed – 0.75v Rd,c ) s r u 1 /1.5f ywd,ef ) Exp. (6.52) where s r = 175 mm Cl. 9.4.3(1) f ywd,ef = effective design strength of reinforcement = (250 + 0.25d) < f yd = 309 MPa Cl. 6.4.5(1) For perimeter u 1 A sw = (1.17 – 0.75 × 0.68) × 175 × 4553/(1.5 × 309) = 1135 mm 2 per perimeter Try 15 no. H10 (1177 mm 2 ) § See Section 3.4.14 with respect to possible limit of 2.0 or 2.5 on V Ed /V Rd,c within punching shear requirements. # v Rd,c for various values of d and r l is available from Table C6. 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 132 Check availability of reinforcement ‡ : 1st perimeter to be > 0.3d but < 0.5d, i.e between 70 mm and 117 mm from face of column. Say 0.4d = 100 mm from face of column. Fig. 9.10, 9.4.3(4) By inspection of Figure 4.21. the equivalent of 14 locations are available between 70 mm and 117 mm from face of column therefore say OK. 150 470 200 200 470 C L C L u 1 s = 150 mm 24 H10 legs in u 1 perimeter H10 legs of links 150 150 150 150 150 900 600 70 58 1 7 5 1 7 5 1 7 5 1 7 5 1 7 5 1 7 5 Figure 4.21 Shear links and punching shear perimeter u 1 Perimeter at which no punching shear links are required: u out = V Ed × b/(d × v Rd,c ) u out = 1087 × 1.15 × 10 3 /(235 × 0.68) = 7826 mm Length of column faces = 4 × 400 = 1600 mm Radius to u out = (7823 – 1600)/2π = 990 mm from face of column i.e. in ribs, therefore beam shear governs ‡ The same area of shear reinforcement is required for all perimeters inside or outside perimeter u 1 . See Section 3.4.13. Punching shear reinforcement is also subject to requirements for minimum reinforcement and spacing of shear reinforcement (see Cl. 9.4.3). The centre of links from the centreline of the column shown in Figure 4.21 have been adjusted to accommodate a perimeter of links at between 0.3d and 0.5d from the column face. This publication is available for purchase from www.concretecentre.com 133 4.3.11 Summary of design A 14H25T H10 links in 12 legs @ 150 cc 13H25B 13H25B 12H20B 12H20B B C D E X X 14H25T 12H25T 12H20T 12H20T Figure 4.23 Section X–X 43 Cont|nuous w|de ¹-|eem This publication is available for purchase from www.concretecentre.com 134 Columns General ¹he ce|cu|et|ons |n th|s sect|on |||ustrete S! Les|gn o¦ e non-s|ender edge co|umn us|ng hend ce|cu|et|on S? Les|gn o¦ e per|meter co|umn us|ng |teret|on o¦ equet|ons to determ|ne re|n¦orcement requ|rements S3 Les|gn o¦ en |nterne| co|umn w|th h|gh ex|e| |oed S4 Les|gn o¦ e s|ender co|umn requ|r|ng e two-hour ure res|stence ln genere|, ex|e| |oeds end urst order moments ere essumed to |e eve||e||e ¹he des|gns cons|der s|enderness |n order to determ|ne des|gn moments, / ld ¹he co|umns ere des|gned end chec|ed ¦or ||ex|e| |end|ng ¹he e¦¦ects o¦ e||ow|ng ¦or |mper¦ect|ons ere |||ustreted A genere| method o¦ des|gn|ng co|umns |s es ¦o||ows ln prect|ce, severe| o¦ these steps mey |e com||ned Leterm|ne des|gn ||¦e N EC0 & NA Table NA 2.1 Assess ect|ons on the co|umn N EC1 (10 parts) & UK NAs Leterm|ne wh|ch com||net|ons o¦ ect|ons epp|y N EC0 & NA Tables NA A1.1 & NA A1.2(B) Assess dure||||ty requ|rements end determ|ne concrete N strength BS 8500–1 Chec| cover requ|rements ¦or eppropr|ete ¦|re N res|stence per|od Approved Document B, EC2–1–2 Leterm|ne cover ¦or ¦|re, dure||||ty end |ond N Cl. 4.4.1 Ane|yse structure ¦or cr|t|ce| com||net|on moments N end ex|e| ¦orces Section 5 Chec| s|enderness end determ|ne des|gn moments N Section 5.8 Leterm|ne eree o¦ re|n¦orcement requ|red N Section 6.1 Chec| spec|ng o¦ |ers end ||n|s N Sections 8 & 9 5 5.0 This publication is available for purchase from www.concretecentre.com Co|umns 135 A 300 mm square column on the edge of a flat slab structure supports an axial load of 1620 kN and first order moments of 38.5 kNm top and −38.5 kNm bottom in one direction only ‡ . The concrete is grade C30/37, f ck f = 30 MPa and cover, c nom , = 25 mm. The 250 mm thick flat slabs are at 4000 mm vertical centres. N Ed = 1620 kN 38.5 kNm – 38.5 kNm Figure 5.1 Forces in edge column 5.1.1 Check slenderness, l Effective length § , l 0 = factor × l Cl. 5.8.3.2 where factor = from Table C16, condition 2 each end Table C16, Table C16, = 0.85 PD 6687: 2.10 l = clear height = 3750 mm l = l 0 = 0.85 × 3750 = 3187 mm Slenderness l = l 0 /i Exp. (5.14) ‡ For examples of load take-downs and 1st order moment analysis see Section 5.3.2 § Effective lengths are covered in Eurocode 2 Cl. 5.8.3.2 and Exp. (5.15). The effective length of most columns will be l /2< l 0 l < l (see Eurocode 2 Figure 5.7f). l PD 6687 [6] Cl. 2.10 suggests that using the procedure outlined in Eurocode 2 (5.8.3.2(3) and 5.8.3.2(5)) leads to similar effective lengths to those tabulated in BS 8110 [7] and reproduced in Table 5.1 of Concise Eurocode 2 [5] and in this publication as Table C16. For simplicity, tabular values are used in this example. However, experience suggests that these tabulated values are conservative. Fig. 5.7 PD 6687 [6] : Cl. 2.10 Cl. 5.8.3.2(3) 5.8.3.2(5) Table C16 Table C16 5.1 Edge column ¹he |ntent|on o¦ th|s ce|cu|et|on |s to show e typ|ce| hend ce|cu|et|on thet me|es re¦erence to des|gn cherts lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Edge column This publication is available for purchase from www.concretecentre.com 136 where i = radius of gyration = h/12 0.5 for rectangular sections l = 3187 × 3.46/300 = 36.8 5.1.2 Limiting slenderness, l lim l lim = 20 ABC/n 0.5 Exp. (5.13N) where A = 0.7 (default) Cl. 5.8.3.1(1) B = 1.1 (default) C = 1.7 − r m = 1.7 − M 01 /M 02 = 1.7 − 38.5/(−38.5) = 2.7 n = N Ed /A c f cd = 1620 × 10 3 /(300 2 × 0.85 × 30/1.5) = 1.06 l lim = 20 ABC/n 0.5 = 20 × 0.7 × 1.1 × 2.7/1.06 0.5 In this example l lim = 40.4 i.e. > 36.8 = Column not slender 5.1.3 Design moments M Ed = max[M 02 ; M 0Ed + M 2 ; M 01 + 0.5M 2 ] Cl. 5.8.8.2(1) where M 02 = M + e i N Ed ≥ e 0 N Ed Cl. 5.8.8.2, 6.1.4 where M = 38.5 kNm e i = l 0 /400 e 0 = max[h/30; 20] = max[300/30; 20] = 20 mm Cl. 5.2.7, 5.2.9 Cl. 6.1.4 M 02 = 38.5 + 1620 × 3.187/400 ≥ 0.02 × 1620 = 38.5 + 12.9 ≥ 32.4 kNm = 51.4 kNm M 0Ed = 0.6M 02 + 0.4M 01 ≥ 0.4M 02 = 0.6 × 51.4 + 0.4 × (−38.5 + 12.9) ≥ 0.4 × 51.4 = 20.6 ≥ 20.6 = 20.6 M 2 = 0 (column is not slender) M 01 = M 02 = max[M 02 ; M 0Ed + M 2 ; M 01 + 0.5M 2 ] = 51.4 kNm = M Ed = 51.4 kNm 5.1.4 Design using charts (see Appendix C) d 2 = c nom + link + f/2 = 25 + 8 + 16 = 49 d 2 /h = 49/300 = 0.163 = interpolating between d 2 /h = 0.15 and 0.20 Figs. C5c), C5d) for N Ed /bhf ck = 1620 × 10 3 /(300 2 × 30) = 0.60 This publication is available for purchase from www.concretecentre.com 137 S! ldge co|umn M Ed /bh 2 f ck = 51.4 × 10 6 /(300 3 × 30) = 0.063 A s f yk /bhf ck = 0.24 A s = 0.24 × 300 2 × 30/500 = 1296 mm 2 Try 4 no. H25 (1964 mm 2 ) 5.1.5 Check for biaxial bending l y /l z ≈ 1.0 Cl. 5.8.9 i.e. l y /l z ≤ 2.0 = OK but check Exp. (5.38b) Exp. (5.38a) As a worst case M Edy may coexist with e 0 N Ed about the orthogonal axis: Cl. 6.1(4) e y /h eq = (M Edz /N Ed )/h = M Edz e z /b eq (M Edy /N Ed )/b M Edy Exp. (5.38b) Imperfections need to be taken into account in one direction only. = As a worst case for biaxial bending Cl. 5.8.9(2) M Edz = M + 0 = 38.5 kNm M Edy = e 0 N Ed = 32.4 kNm M Edz = 38.5 = 1.19 i.e. > 0.2 and < 5.0 M Edy 32.4 Exp. (5.38b) = Biaxial check required Cl. 5.8.9(4) Check whether (M Edz /M Rdz ) a + (M Edy /M Rdy ) a ≤ 1.0 Exp. (5.39) where M Edz = 38.5 kNm M Edy = 32.4 kNm M Rdz = M Rdy Figs. C5c), C5d) To determine M Rdz , find M Ed /bh 2 f ck (and therefore moment capacity) by interpolating between d 2 /h = 0.15 (Figure C5c) and 0.20 (Figure C5d) for the proposed arrangement and co-existent axial load. Assuming 4 no. H25, A s f yk /bhf ck = 1964 × 500/(300 2 × 30) = 0.36 Interpolating for N Ed /bhf ck = 0.6, M Ed /bh 2 f ck = 0.094 = M Rdz = M Rdy = 0.094 × 300 3 × 30 = 76.1 kNm a is dependent on N Ed /N Rd where N Ed = 1620 kN as before Cl. 5.8.9(4), Notes to Exp. (5.39) This publication is available for purchase from www.concretecentre.com 138 N Rd = A c f cd + A s f yd = 300 2 × 0.85 × 30/1.5 + 1964 × 500/1.15 = 1530.0 + 853.9 = 2383.9 kN N Ed /N Rd = 1620/2383.9 = 0.68 =a = 1.48 by interpolating between values given for N Ed /N Rd = 0.1, (1.0) and N Ed /N Rd = 0.7, (1.5) (M Edz /M Rdz ) a + (M Edy /M Rdy ) a = (38.5/76.1) 1.48 + (32.4/76.1) 1.48 = 0.36 + 0.28 = 0.64 = OK. = 4 no. H25 OK Exp. (5.39) 5.1.6 Links Diameter min. f/4 = 25/4 = 8 mm Cl. 9.5.3 & NA Max. spacing = 0.6 × 300 = 180 mm Cl. 9.5.3(3), Cl. 9.5.3(4) Links at say 175 mm cc 5.1.7 Design summary 4 H25 H8 links @ 175 cc 25 mm cover f ck = 30 MPa Figure 5.2 Design summary: edge column This publication is available for purchase from www.concretecentre.com 139 S? ler|meter co|umn This 300 × 300 mm perimeter column is in an internal environment and supports three suspended floors and the roof of an office block. It is to be designed at ground floor level where the storey height is 3.45 m and the clear height in the N–S direction (z direction) is 3.0 m and 3.325 m in the E–W direction (y direction). One-hour fire resistance is required and f ck = 30 MPa. 250 325 300 × 300 Column under consideration 3575 M 02yy Figure 5.3 Perimeter column (internal environment) From first order analysis, load case 1: N Ed = 1129.6 kN; M 02yy = 89.6 kNm; M 02zz = 0 Load case 2: N Ed = 1072.1 kN; M 02yy = 68.7 kNm; M 02zz = 6.0 kNm 5.2.1 Cover c nom = c min + Dc dev Exp. (4.1) where c min = max[c min,b ; c min,dur ] where c min,b = diameter of bar. Assume 32 mm bars and 8 mm links Cl. 4.4.1.2(3) = 32 mm to main bars, 32 − 8 = 24 mm to links = say 25 mm c min,dur = minimum cover due to environmental conditions. Assume XC1. c min,dur = 15 mm c min = 25 mm Dc dev = 10 mm Cl. 4.4.1.3(3) Therefore c nom = 25 + 10 = 35 mm to links c nom = 35 mm to links. lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Perimeter column (internal environment) 5.2 Perimeter column (internal environment) ¹h|s exemp|e |s |ntended to show e hend ce|cu|et|on ¦or e non-s|ender per|meter co|umn us|ng |teret|on (o¦ \) to determ|ne the re|n¦orcement requ|red This publication is available for purchase from www.concretecentre.com 140 5.2.2 Fire resistance Check validity of using Method A and Table 5.2a of BS EN 1992–1–2: EC2-1-2: 5.3.2, Table 5.2a l 0,fi ≈ 0.7 × 3.325 i.e. < 3.0 m = OK. EC2-1-2: 5.3.3(3) e = M 02yy /N Ed = 89.6 × 10 6 /1129.6 × 10 3 = 79 mm EC2-1-2: 5.3.2 & NA e max = 0.15h = 0.15 × 300 = 45 mm = no good. Check validity of using Method B and Table 5.2b: e max = 0.25b = 75 mm = no good. EC2-1-2: 5.3.3 Use BS EN 1992–1–2 Annex C Tables C1–C9. EC2-1-2: Annex C Assume min. 4 no. H25 = 1964 mm 2 (z 2.2%) ‡ w = A s f yd /A c f cd = 0.022 × (500/1.15)/(0.85 × 30/1.5) EC2-1-2: 5.3.3(2) = 0.56 e ≈ 0.25b and ≤ 100 mm l = l 0 /i where l 0 = 0.7 × 3.325 = 2327 mm i = radius of gyration = (I/A) 0.5 = h/12 0.5 EC2-1-2: 5.3.3(2), 5.3.3(3) where I = inertia = bh 3 /12 A = area = bh h = height of section b = breadth of section = 300/12 0.5 = 87 mm l = 2327/87 = 276 n = N 0Ed,fi /0.7(A c f cd + A s f yd ) EC2-1-2: 5.3.3(2) = 0.7 × 1129.6/0.7(300 2 × 0.85 × 30/1.5 + 1964 × 500/1.15) = 1129.6/2383.9 = 0.47 = interpolate for l = 30 and n = 0.47 between from Table C.5 of BS EN 1992–1–2 (w = 0.5, e = 0.25b): EC2-1-2: Table C.5 minimum dimension, b min = 235, and axis distance, a = 35 mm and from Table C.8 of BS EN 1992–1–2 (w = 1.0, e = 0.25b): EC2-1-2: Table C.8 ‡ Using 4 no. H20 gives w = 0.34, n = 0.54 and b min = 310 mm = no good. This publication is available for purchase from www.concretecentre.com 141 S? ler|meter co|umn b min = 185, and a = 30 mm = for w = 0.56, b min = 228, and a = 35 mm OK to use Method B but use min. 4 no. H25 5.2.3 Structural design: check slenderness Effective length, l 0 : l 0 = 0.5l [1 + k 1 /(0.45 + k 1 ) ] 0.5 [1 + k 2 /(0.45 + k 2 ) ] 0.5 Exp. (5.15) where k 1 , k 2 = relative stiffnesses top and bottom But conservatively, choose to use tabular method § . For critical direction, the column is in condition 2 at top and condition 3 at bottom (pinned support). l 0 = 0.95 × 3325 = 3158 mm Table C16 Slenderness ratio, l: l = l 0 /i where i = radius of gyration = (I/A) 0.5 = h/12 0.5 l = 3158 × 12 0.5 /300 = 36.5 l = 36.5 Cl. 5.8.3.2(1) Limiting slenderness ratio, l lim l lim = 20 ABC/n 0.5 Cl. 5.8.3.1(1) & NA where A = 1/(1 + 0.2 f ef ). Assume 0.7 Cl. 5.8.4 B = (1 + 2 A s f yd /A c f cd ) 0.5 = (1 + 2w) 0.5 Assuming min. 4 no. H25 (for fire) w = 0.56 as before B = (1 + 2 × 0.56) 0.5 = 1.46 Cl. 5.8.3.1(1) C = 1.7 – r m Cl. 5.8.3.1(1) where r m = M 01 /M 2 Assuming conservatively that M 01 = 0 r m = 0 C = 1.7 n = N Ed /A c f cd = 1129.6 × 10 3 /(300 2 × 0.85 × 30/1.5) = 0.74 § See footnote to Section 5.1.1. This publication is available for purchase from www.concretecentre.com 142 l lim = 20 × 0.7 × 1.46 × 1.7/0.74 0.5 = 40.4 l lim = 40.4 = as l < l lim column is not slender and 2nd order moments are not required. Column is not slender 5.2.4 Design moments, M Ed M Ed = M OEd + M 2 ≥ e 0 N Ed Cl. 5.8.8.2(1), 5.8.8.2(3) But as column is not slender, M 2 = 0, = Cl. 6.1.4 M Ed = M Oed = M + e i N Ed ≥ e 0 N Ed where M = moment from 1st order analysis e i N Ed = effect of imperfections ‡ where e i = l 0 /400 Cl. 5.2(7), 5.2.9, 5.8.8.2(1) e 0 = h/30 > 20 mm Cl. 6.1.4 Load case 1: M Edy = 89.6 + (3158/400) × 1129.6 × 10 −3 > 0.02 × 1129.6 = 89.6 + 8.9 > 22.6 = 98.5 kNm Load case 2: M Edy = 68.7 kNm M Edz = 6.0 + (l 0 /400) × 1072.1 × 10 −3 > 0.02 × 1072.1 where l 0 = 0.9 × 3000 = 13.2 > 21.4 = 21.4 kNm Table C16 5.2.5 Design using iteration of x For axial load: A sN /2 = (N Ed – a cc nf ck bd c /g C )/(s sc – s st ) Concise: Section 6.2.2, Appendix A3 For moment: A sM /2 = [M Ed – a cc nf ck bd c (h/2 – d c /2)/ g C ] (h/2 – d 2 ) (s sc – s st ) Appendices A3, C9.2, where M Ed = 98.5 × 10 6 N Ed = 1129.6 × 10 3 a cc = 0.85 Cl. 3.1.6(1) & NA n = 1.0 for f ck ≤ 50 MPa Exp. (3.21) ‡ The effects of imperfections need only be taken into account in the most unfavourable direction. Cl. 5.8.9(2) This publication is available for purchase from www.concretecentre.com 143 S? ler|meter co|umn f ck = 30 b = 300 h = 300 d c = depth of compression zone = lx Exp. (3.19) = 0.8x < h where x = depth to neutral axis d 2 = 35 + 8 + 25/2 = 55 mm assuming H25 g C = 1.5 s sc , (s st ) = stress in reinforcement in compression (tension) Table 2.1N a) Strain diagram b) Stress diagram s st f cd = a cc nf ck /g C d 2 h x d c d 2 A s2 A s1 o o o o n. axis s sc e sc e cu2 e y Figure 5.4 Section in axial compression and bending Fig. 6.1 Try x = 200 mm e cu = e cu2 = 0.0035 e sc = 0.0035 × (x – d 2 ) = 0.0035 × (200 – 55) x 200 = 0.0025 s sc = 0.0025 × 200000 ≤ f yk /g S = 500 ≤ 500/1.15 = 434.8 MPa e st = 0.0035(h – x – d 2 )/x = 0.0035(300 – 200 – 55)/200 = 0.0008 s st = 0.0008 × 200000 ≤ 500/1.15 = 160 MPa A sN /2 = 1129.6 × 10 3 – 0.85 × 1.0 × 30 × 300 × 200 × 0.8/(1.5 × 10 3 ) 434.8 – 160 = (1129.6 – 816.0) × 10 3 = 1141 mm 2 274.8 This publication is available for purchase from www.concretecentre.com 144 A sM /2 = 98.5 × 10 6 – 0.85 × 1.0 × 30 × 300 × 200 × 0.8 (300/2 – 200 × 0.8/2)/(1.5 × 10 3 ) (300/2 – 55) (434.8 + 160) = (98.5 – 57.1) × 10 6 = 733 mm 2 95 × 594.8 Similarly for x = 210 mm e cu = 0.0035 e sc = 0.0026 = s sc = 434.8 e st = 0.0006 = s st = 120 MPa A sN /2 = (1129.6 – 856.8) × 10 3 = 866 mm 2 434.8 – 120 A sM /2 = (98.5 – 56.5) × 10 6 = 796 mm 2 95 × 554.8 Similarly for x = 212 mm s sc = 434.8 e st = 0.00054 =e st = 109 MPa A sN /2 = (1129.6 – 865.0) × 10 3 = 812 mm 2 434.8 – 109 A sM /2 = (98.5 – 56.3) × 10 6 = 816 mm 2 95 × 543.8 = as A sN /2 ≈ A sM /2, x = 212 mm is approximately correct and A sN ≈ A sM , ≈ 1628 mm 2 = Try 4 no. H25 (1964 mm 2 ) 5.2.6 Check for biaxial bending By inspection, not critical. Cl. 5.8.9(3) [Proof: Section is symmetrical and M Rdz > 98.5 kNm. Assuming e y /e z > 0.2 and biaxial bending is critical, and assuming exponent a = 1 as a worst case for load case 2: Exp. (5.39) (M Edz /M Rdz ) a + (M Edy /M Rdy ) a = (21.4/98.5) 1 + (68.7/98.5) 1 = 0.91 i.e. < 1.0 = OK.] 5.2.7 Links Minimum size links = 25/4 = 6.25, say 8 mm Spacing: minimum of a) 0.6 × 20 × 25 = 300 mm, b) 0.6 × 300 = 180 mm or c) 0.6 × 400 = 240 mm Use H8 @ 175 mm cc Cl. 9.5.3(3), 9.5.3(4) This publication is available for purchase from www.concretecentre.com 145 S? ler|meter co|umn 5.2.8 Design summary 4 H25 H8 links @ 175 cc c nom = 35 mm to links Figure 5.5 Design summary: perimeter column This publication is available for purchase from www.concretecentre.com 146 The design forces need to be determined. This will include the judgement of whether to use Exp. (6.10) or the worse case of Exp. (6.10a) and (6.10b) for the design of this column. The suspended slabs (including the ground floor slab) are 300 mm thick flat slabs at 4500 mm vertical centres. Between ground and 5th floors the columns at C2 are 500 mm square; above 5th floor they are 465 mm circular. Assume an internal environment, 1-hour fire resistance and f ck = 50 MPa. E D C A 4.0 8.0 9.6 200 x 200 hole 200 x 200 hole 300 mm flat slabs All columns 400 mm sq. 8.6 8.0 4.0 4.0 6 .0 1 2 3 B Bb 5.3.1 Design forces In order to determine design forces for this column it is first necessary to determine vertical loads and 1st order moments. 5.3.2 Load take-down Actions: Roof: g k = 8.5, q k = 0.6 EC1-1-1: 6.3.4, NA & Table NA.7 5.3 Internal column ¹he ¦et s|e| shown |n lxemp|e 34 (reproduced es l|gure S6) |s pert o¦ en S-storey structure e|ove ground w|th e |esement |e|ow ground ¹he pro||em |s to des|gn co|umn C? |etween ground ¦oor end !st ¦oor lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Internal column Figure 5.6 Part plan of flat slab This publication is available for purchase from www.concretecentre.com 147 S3 lnterne| co|umn Floors: g k = 8.5, q k = 4.0 Section 3.4 In keeping with Section 3.4 use coefficients to determine loads in take-down. Section 3.4 Consider spans adjacent to column C2: Along grid C, consider spans to be 9.6 m and 8.6 m and C2 to be the internal of 2 - span element. Therefore elastic reaction factor = 0.63 + 0.63 = 1.26 Along grid 2 consider spans to be 6.0 m and 6.2 m and internal of multiple span. Elastic reaction factor = 0.5 + 0.5 = 1.00 Table C3 Load take-down for column C2. Item Calculation G k Q k From item Cumulative total From item Cumulative total Roof = [erf y × (l z1 + l z2 )/2 ] × [erf z × (l y1 + l y2 )/2 ] × (g k + q k ) = [1.0 × (6.0 + 6.2)/2 ] × [1.26 × (9.6 + 8.6)/2 ] × (8.5 + 0.6) = 69.9 × (8.5 + 0.6) = 594.5 42.0 Col 8 – R = π (0.465/2) 2 × (4.5 − 0.3) × 25 = 17.9 612.4 42.0 8th = 1.0 × (6.0 + 6.2)/2 × 1.26 × (9.6 + 8.6)/2 × (8.5 + 4.0) = 594.5 279.7 Col 7 – 8 as before 17.9 1224.8 321.7 7th a.b. 594.5 279.7 Col 6 – 7 a.b. 17.9 1837.2 601.4 6th a.b. 594.5 279.7 Col 5 – 6 a.b. 17.9 2449.6 881.1 5th a.b. 594.5 279.7 Col 4 – 5 = 0.5 × 0.5 × (4.5 − 0.3) × 25 = 26.3 3070.4 1160.8 4th as before 594.5 279.7 Col 3 – 4 a.b. 26.3 3691.2 1440.5 3rd a.b. 594.5 279.7 Col 2 – 3 a.b. 26.3 4312.0 1720.2 2nd a.b. 594.5 279.7 Col 1 – 2 a.b. 26.3 4932.8 1999.7 1st a.b. 594.5 279.8 Col G – 1 a.b. 26.3 5553.6 2279.5 At above ground floor ÷ 5553.6 ÷ 2279.5 This publication is available for purchase from www.concretecentre.com 148 5.3.3 Design axial load, ground– 1st floor, N Ed a) Axial load to Exp. (6.10) N Ed = g G G k + g Q Q k1 + c 0 g Q Q ki EC0: Exp. (6.10) & NA where g G = 1.35 g Q = 1.50 c 0,1 = 0.7 (offices) Q k1 = leading variable action (subject to reduction factor a A or a n ) EC0: A1.2.2 & NA Q ki = accompanying action (subject to a A or a n ) EC1-1-1: 6.3.1.2 (10), 6.3.1.2 (11), & NA where a A = 1 – A/1000 ≥ 0.75 = 1 – 9 × 69.9/1000 = 0.37 ≥ 0.75 = 0.75 a n = 1.1 – n/10 for 1 ≤ n ≤ 5 = 0.6 for 5 ≤ n ≤ 10 and = 0.5 for n > 10 where n = number of storeys supported a n = 0.6 for 8 ‡ storeys supported = as a n < a A , use a n = 0.6 Assuming the variable action of the roof is an independent variable action: N Ed = 1.35 × 5553.6 + 1.5 × (2279.5 − 42.0) × 0.6 + 0.7 × 1.5 × 42.0 = 1.35 × 5553.6 + 1.5 × 2237.5 + 0.7 × 1.5 × 42.0 = 7497.4 + 2013.8 + 44.1 = 9555.3 kN To Exp. (6.10), N Ed = 9555.3 kN b) Axial load to Exp. (6.10a) N Ed = g G G k + c 0,1 g Q Q k1 + c 0,1 g Q Q ki = 1.35 × 5553.6 + 0.7 × 1.5 × 0.6 ( 279.8 + 1999.7) = 7497.4 + 1436.1 = 8933.4 kN To Exp. (6.10a), N Ed = 8933.4 kN EC0: Exp. (6.10a) & NA c) Axial load to Exp. (6.10b) N Ed = jg G G k + g Q Q k1 + c 0,1 g Q Q ki EC0: Exp. (6.10) & NA ‡ According to BS EN 1991–1–1 6.3.1.2 [11] the imposed load on the roof is category H and therefore does not qualify for reduction factor a n . This publication is available for purchase from www.concretecentre.com 149 S3 lnterne| co|umn assuming the variable action of the roof is an independent variable action: = 0.925 × 1.35 × 5553.6 + 1.5 × (2279.5 − 42.0) × 0.6 + 0.7 × 1.5 × 42.0 = 1.25 × 5553.6 + 1.5 × 2237.5 × 0.6 + 0.7 × 1.5 × 42.0 = 6942.1 + 2013.8+ 44.1 = 9000.0 kN To Exp. (6.10b), N Ed = 9000.0 kN 5.3.4 First order design moments, M a) Grid C Consider grid C to determine M yy in column (about grid 2) 300 thick ave 6100 wide 500 sq q k = 6.1 x 4.0 = 24.4 kN/m g k = 6.1 x 8.5 = 51.9 kN/m 4500 4500 1 2 3 8600 9600 Actions: g k = (6.0 + 6.2)/2 × 8.5 = 51.9 kN/m q k = (6.0 + 6.2)/2 × 4.0 = 24.4 kN/m Relative stiffness of lower column: Assuming remote ends of slabs are pinned, relative stiffness = b lc d lc 3 /L lc b lc d lc 3 /L lc + b uc d uc 3 /L uc + 0.75b 23 d 23 3 /L 23 + 0.75b 21 d 21 3 /L 21 where b = breadth d = depth L = length lc = lower column, uc = upper column 23 = beam 23, similarly 21 = beam 21 = 0.5 4 /4.5 2 × 0.5 4 /4.5 + 0.75 × 6.1 × 0.3 3 /8.6 + 0.75 × 6.1 × 0.3 3 /9.6 = 0.0139/(0.0278 + 0.0144 + 0.0129) = 0.252 This publication is available for purchase from www.concretecentre.com 150 1st order moment using Exp. (6.10) FEM 23 ‡ = 1.35 × 51.9 × 8.6 2 /12 = 431.8 kNm FEM 21 = (1.35 × 51.9 + 1.5 × 24.4) × 9.6 2 /12 = 106.7 × 9.6 2 /12 = 819.5 kNm M lower,yy = 0.252 × [819.5 – 431.8] = 97.7 kNm 1st order moment using Exp. (6.10a) FEM 23 = 1.25 × 51.9 × 8.6 2 /12 = 399.8 kNm FEM 21 = (1.25 × 51.9 + 1.5 × 24.4) × 9.6 2 /12 = 101.5 × 9.6 2 /12 = 779.5 kNm M lower,yy = 0.252 × (779.5 – 399.8) = 95.7 kNm 1st order moment using Exp. (6.10b) FEM 23 = 1.35 × 51.9 × 8.6 2 /12 = 431.8 kNm FEM 21 = (1.35 × 51.9 + 0.7 × 1.5 × 24.4 ) × 9.6 2 /12 = 95.7 × 9.6 2 /12 = 735.0 kNm M lower,yy = 0.252 × (735.0 – 431.8) = 76.4 kNm = Exp. (6.10a) critical b) Grid 2 Consider grid 2 to determine M zz in column (about grid C) 300 thick ave 11470 wide 500 sq g k = 97.5 kN/m q k = 45.9 kN/m 4500 4500 D C B 6200 6000 Figure 5.8 Subframe on column C2 along grid 2 Actions: g k = 0.63 × (8.6 + 9.6) × 8.5 = 11.47 × 8.5 = 97.5 kN/m q k = 11.47 × 4.0 = 45.9 kN/m Relative stiffness of lower column: Assuming remote ends of slabs are fixed, relative stiffness Cl. 5.8.3.2(4) PD 6687 ‡ FEM 23 = Fixed end moment in span 23 at grid 2. This publication is available for purchase from www.concretecentre.com 151 S3 lnterne| co|umn = 0.5 4 /4.5 2 × 0.5 4 /4.5 + 11.47 × 0.3 3 /6.2 +11.47 × 0.3 3 /6.0 = 0.0139/(0.0278 + 0.0500 + 0.0516) = 0.107 1st order moment using Exp. (6.10) FEM CB = (1.35 × 97.5 + 1.5 × 45.9) × 6.2 2 /12 = 200.5 × 6.2 2 /12 = 642.3 kNm FEM CD = 1.35 × 97.5 × 6.0 2 /12 = 394.9 kNm M lower,zz = 0.107 × (642.3 – 394.9) = 26.5 kNm 1st order moment using Exp. (6.10a) FEM CB = 1.25 × 97.5 × 6.0 2 /12 = 365.6 kNm FEM CD = (1.25 × 97.5 + 1.5 × 45.9) × 6.22/12 = 190.7 × 6.22/12 = 611.0 kNm M lower,zz = 0.107 × (611.0 – 365.6) = 26.3 kNm 1st order moment using Exp. (6.10b) FEM CB = (1.35 × 97.5 + 0.7 × 1.5 × 45.9) × 6.2 2 /12 = 190.7 × 6.2 2 /12 = 576.0 kNm FEM CD = 1.35 × 97.5 × 6.0 2 /12 = 394.9 kNm M lower,zz = 0.107 × (576.0 – 394.9) = 19.4 kNm = Exp. (6.10a) critical again 5.3.5 Summary of design forces in column C2 ground–1st Design forces Method N Ed M yy about grid 2 M zz about grid C Using Exp. (6.10) 9555.3 kN 97.7 kNm 26.5 kNm Using Exp. (6.10a) 8933.4 kN 95.7 kNm 26.3 kNm Using Exp. (6.10b) 9000.0 kN 76.4 kNm 19.4 kNm Notes: 1) To determine maximum 1st order moments in the column, maximum out-of- balance moments have been determined using variable actions to one side of the column only. The effect on axial load has, conservatively, been ignored. 2) It may be argued that using coefficients for the design of the slab and reactions to the columns does not warrant the sophistication of using Exps (6.10a) and (6.10b). Nevertheless, there would appear to be some economy in designing the column to Exp. (6.10a) or Exp. (6.10b) rather than Exp. (6.10). The use of Exp. (6.10a) or Exp. (6.10b) is perfectly valid and will be followed here. To avoid duplicate designs for both Exps (6.10a) and (6.10b), a worse case of their design forces will be used, thus: N Ed = 9000 kN, M yy = 95.7 kNm, M zz = 26.3 kNm This publication is available for purchase from www.concretecentre.com 152 5.3.6 Design: cover c nom = c min + Dc dev Exp. (4.1) where c min = max[c min,b ; c min,dur ] where c min,b = diameter of bar. Assume 32 mm bars and 8 mm links. = 32 – 8 = 24 mm to link c min,dur = minimum cover due to environmental conditions. Assume XC1. Cl. 4.4.1.2(3) c min,dur = 15 mm BS 8500-1: Table A4 c min = 24 mm, say 25 mm to link Dc dev = 10 mm = c nom = 25 + 10 = 35 mm Cl. 4.4.1.3 & NA 5.3.7 Design: fire resistance Check validity of using Method A and Table 5.2a EC2-1-2: 5.3.2, Table 5.2a a) Check l 0,fi ≤ 3.0 m where l 0 = effective length of column in fire = 0.5 × clear height = 0.5 × (4500 – 300) = 2100 mm OK b) Check e ≤ e max = 0.15h = 0.15 × 500 = 75 mm e = M 0Ed,fi /N 0Ed,fi = M 0 /N Ed = 99.5 × 10 6 /8933 × 10 3 = 11 mm OK EC2-1-2: 5.3.2(2) c) Check amount of reinforcement ≤ 4% OK Assuming m fi = 0.7 b min = 350 with a min = 40 mm OK For fire using Method A and Table 5.2a is valid EC2-1-2: Table 5.2a 5.3.8 Structural design: check slenderness Effective length, l 0 : l 0 = 0.5l [1 + k 1 /(0.45 + k 1 )] 0.5 [1 + k 2 /(0.45 + k 2 )] 0.5 Exp. (5.15) where k 1 and k 2 are relative flexibilities at top and bottom of the column. k i = (EI col /l col )/S(2EI beam /l beam ) ≥ 0.1 PD 6687 [6]‡ ‡ PD 6687 states that to allow for cracking, the contribution of each beam should be taken as 2EI/l beam This publication is available for purchase from www.concretecentre.com 153 S3 lnterne| co|umn Critical direction is where k 1 and k 2 are greatest i.e. where slab spans are greater k 1 = k 2 = b lc d lc 3 /L lc 2b 23 d 23 3 /L 23 + 2b 21 d 21 3 /L 21 = (0.5 4 /4.5)/(2 × 6.1 × 0.3 3 /8.6 + 2 × 6.1 × 0.3 3 /9.6) = (0.0625)/(0.0383 + 0.0343) = 0.86 How to [8] : Columns l 0 = 0.5 (4500 – 300) [1 + 0.86/(0.45 + 0.86)] 0.5 [1 + 0.86/(0.45 + 0.86)] 0.5 l 0 = 0.5 × 4200 × 1.66 = 0.828 × 4200 = 3478 mm Slenderness ratio, l: Cl. 5.8.3.2(1) l = l 0 /i where i = radius of gyration = (I/A) 0.5 = h/12 0.5 = l = 3478 × 12 0.5 /500 = 24.1 Limiting slenderness ratio, l lim : Cl. 5.8.3.1(1) & NA l lim = 20 ABC/n 0.5 where A = 1/(1 + 0.2 f ef ). Assume 0.7 as per default B = (1 + 200) 0.5 . Assume 1.1 as per default C = 1.7 – · m where · m = M 01 /M 2 = –84.9/109.3 = –0.78 C = 1.7 + 0.78 = 2.48 n = N Ed /A c f cd = 8933 × 10 3 /(500 2 × 0.85 × 50/1.5) = 1.26 = l lim = 20 × 0.7 × 1.1 × 2.48/1.26 0.5 = 34.0 = as l < l lim column is not slender and 2nd order moments are not required Exp. (5.13N) 5.3.9 Design moments, M Ed M Ed = M + e i N Ed ≥ e 0 N Ed Cl. 5.8.8.2(1), 6.1(4) where M = moment from 1st order analysis e i N Ed = effect of imperfections Cl. 5.8.8.2(1) This publication is available for purchase from www.concretecentre.com 154 where e i = l 0 /400 Cl. 5.2.7 e 0 N Ed = minimum eccentricity Cl. 6.1(4) where e 0 = h/30 ≥ 20 mm M Edyy = 95.7 + (3570/400) × 8933 × 10 −3 ≥ 0.02 × 8933 = 95.7 + 79.7 ≥ 178.7 = 175.4 < 178.7 kNm M Edzz = 18.8 + 79.7 ≥ 178.7 = 178.7 kNm = Both critical. However, imperfections need only be taken in one direction – where they have the most unfavourable effect = Use M Edzz = 178.7 with M Edyy = 95.7 kNm Cl. 5.8.9(2) 5.3.10 Design using charts M Edyy /bh 2 f ck = 178.9 × 10 6 /(500 3 × 50) = 0.03 N Ed /bhf ck = 9000 × 10 3 /(500 2 × 50) = 0.72 Choice of chart based on d 2 /h where d 2 = depth to centroid of reinforcement in half section assuming 12 bar arrangement with H32s d 2 = 35 + 8 + (32/2) + (2/6) [500 + 2 × (35 + 8 + 32/2 )/3] = 59 + (1/3) × 127 = 101 = d 2 /h = 101/500 = 0.2 Use Figure C5d) Figs C5a) to C5e) d 2 C L d 2 , to centroid of reinforcement in half section From Figure C5d) Fig. C5d) A s f yk /bhf ck = 0.30 A s = 0.29 × 500 × 500 × 50/500 = 7500 mm 2 Try 12 no. H32 (9648 mm 2 ) ‡ ‡ Using design actions to Exp. (6.10) would have resulted in a requirement for 8500 mm 2 . This publication is available for purchase from www.concretecentre.com 155 S3 lnterne| co|umn 5.3.11 Check biaxial bending Cl. 5.8.9 Slenderness: l y ≈ l z = OK. Eccentricities: as h = b check e y /e z M Edz critical. (Imperfections act in z direction.) e y /e z = 95.7 × 10 6 /9000 × 10 3 178.7 × 10 6 /9000 × 10 3 Cl. 5.8.9(3) = 0.54 i.e. > 0.2 and < 5 = Design for biaxial bending. Cl. 5.9.3(3), Exp. (5.38b) e y e z M Edy Centre of reaction M Edz b h z y * C 2 Figure 5.10 Eccentricities 5.3.12 Design for biaxial bending Check (M Edz /M Rdz ) a + (M Edy /M Rdy ) a ≤ 1.0 Cl. 5.9.3(4), Exp. (5.39) For load case 2 where M Edz = 178.7 kNm M Edy = 95.7 kNm M Rdz = M Rdy = moment resistance. Using charts: From Figure C4d), for d 2 /h = 0.20 and A s f yk /bhf ck = 9648 × 500/500 × 500 × 50 = 0.39 N Ed /bhf ck = 9000 x 10 3 /(500 2 x 50) = 0.72 M Rd /bh 2 f ck = 0.057 Fig. C5d) = M Rd ≈ 0.057 × 500 3 × 50 = 356.3 kNm This publication is available for purchase from www.concretecentre.com 156 a = exponent dependent upon N Ed /N Rd Cl. 5.8.3(4) where N Rd = A c f cd + A s f yd = 500 × 500 × 0.85 × 50/1.5 + 9648 × 500/1.15 = 7083 + 3216 = 10299 kN N Ed /N Rd = 9000/10299 = 0.87. Interpolating between values given for N Ed /N Rd = 0.7 (1.5) and for N Ed /N Rd = 1.0 (2.0) Notes to Exp. (5.39) = a = 1.67 Check (M Edz /M Rdz ) a + (M Edy /M Rdy ) a ≤ 1.0 (178.7/356.3) 1.67 + (95.7/356.3) 1.67 = 0.32 + 0.11 = 0.43 i.e. < 1.0 = OK Use 12 no. H32 5.3.13 Links Minimum diameter of links: = f/4 = 32/4 = 8 mm Cl. 9.5.3 & NA Spacing, either: a) 0.6 × 20 × f = 12 × 32 = 384 mm, b) 0.6 × h = 0.6 × 500 = 300 mm or c) 0.6 × 400 = 240 mm. = Use H8 links at 225 mm cc Cl. 9.5.3(3), 9.5.3(4) Number of legs: Bars at 127 mm cc i.e. < 150 mm = no need to restrain bars in face but good practice suggests alternate bars should be restrained. = Use single leg on face bars both ways @ 225 mm cc Cl. 9.5.3(6) SMDSC: 6.4.2 5.3.14 Design summary 12 H32 H8 links @ 225 cc 35 mm to link 500 mm sq f ck = 50 MPa Figure 5.11 Design summary: internal column This publication is available for purchase from www.concretecentre.com 157 S4 Sme|| per|meter co|umn 5.4 The middle column, B, in Figure 4.5, supports two levels of storage loads and is subject to an ultimate axial load of 1824.1 kN ‡ . From analysis it has moments of 114.5 kNm in the plane of the beam and 146.1 kNm perpendicular to the beam (i.e. about the z axis). The column is 350 mm square, 4000 mm long, measured from top of foundation to centre of slab. It is supporting storage loads, in an external environment (but not subject to de-icing salts) and is subject to a 2-hour fire resistance requirement on three exposed sides. Assume the base is pinned. Slab 300 450 4000 350 x 350 column Paving Foundation Figure 5.12 Perimeter column 5.4.1 Cover Nominal cover, c nom c nom = c min + Dc dev where c min = max[c min,b ; c min,dur ] Exp. (4.1) ‡ G k = 562.1; Q k = 789.1; as column supports loads from 2 levels a n = 0.9; as imposed loads are from storage c 0 = 1.0 =g Q = 1.50 and g Q = 1.35. = Ultimate axial load, N Ed = 1.35 × 562.1 + 1.5 × 0.9 × 789.1 = 1824.1 kN. lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Small perimeter column subject to two-hour fire resistance Small perimeter column subject to two-hour fire resistance ¹h|s ce|cu|et|on |s |ntended to show e sme|| s|ender co|umn su||ect to e requ|rement ¦or ?-hour ure res|stence lt |s |esed on the exemp|e shown |n Sect|on 4? This publication is available for purchase from www.concretecentre.com 158 where c min,b = diameter of bar. Assume 32 mm main bars and 10 mm links c min,dur = minimum cover due to environmental conditions. Assuming primarily XC3/XC4, secondarily XF1, c min,dur = 25 mm Dc dev = allowance in design for deviation = 10 mm = Try c nom = 32 + 10 = 42 mm to main bars or = 25 + 10 = 35 mm to 8 mm links Try c nom = 35 mm to 8 mm links. Cl. 4.4.1.2(3) BS 8500-1 [14] : Table A4 5.4.2 Fire resistance a) Check adequacy of section for R120 to Method A Axis distance available = 43 mm + f/2 Required axis distance to main bars, a for 350 mm square column EC1-1-2: 5.3.1(1) & For m fi = 0.5, a = 45 mm; and NA 5.3.2, Table 5.2a for m fi = 0.7, a = 57 mm, providing: 8 bars used – OK but check later t l t 0,fi ≤ 3 m – OK but check e t ≤ e max = 0.15h = 0.15 × 350 = 52 mm but e = M 0Ed,fi /N 0Ed,fi = 0.7 × 146.1 × 10 6 /0.7 × 1824.1 × 10 3 = 80 mm = no good Try Method B b) Check adequacy of section for R120 to Method B EC2-1-2: 5.3.3, Determine parameters n, w, and e, and check l fi . Assume 4 no. H32 + 4 no. H25 = (5180 mm 2 : 4.2%) Table 5.2b (say 4.2% OK – integrity OK) Cl. 9.5.2(3) n = N 0Ed,fi /0.7(A c f cd + A s f yd ) EC2-1-2: Exp. (5.8a) = 0.7 × 1824.1 × 10 3 /0.7 (350 × 350 × a cc × f ck / g C + 5180 × 500/g S ) = 1276.9 × 10 3 /0.7 (350 × 350 × 0.85 × 30/1.5 + 5180 × 500/1.15) = 1276.9 × 10 3 /0.7 (2082.5 + 2252.0) = 0.42 OK w = mechanical ratio EC2-1-2: 5.3.3(2) = A s f yd /A c f cd ≤ 1.0 = 2252/2082 = 1.08 ≥ 1 But say within acceptable engineering tolerance =use w = 1.0 OK e = first order eccentricity EC2-1-2: Exp. = M 0Ed,fi /N 0Ed,fi (5.8b) This publication is available for purchase from www.concretecentre.com 159 S4 Sme|| per|meter co|umn = 0.7 × 146.1 × 10 6 /0.7 × 1824.1 × 10 3 EC2-1-2: 2.4.2(3) = 80 mm as before z 0.23h. OK l fi = slenderness in fire = l 0,fi /i where l 0,fi = effective length of column in fire EC2-1-2: 5.3.2(2) = 0.7l = 0.7 × 4000 = 2800 mm i = radius of gyration = h/3.46 for a rectangular section Note 2 = l fi = 2800/(350/3.46) = 27.7 < 30 = OK Table 5.2b valid for use in this case. Interpolating from BS EN 1992–1–2 Table 5.2b for n = 0.42 and w = 1.0, column width = 350 mm and axis distance = say, 48 mm = Axis distance = 43 mm + f/2 is OK c) As additional check, check adequacy of section to Annex B3 and Annex C Using BS EN 1992–1–2 Table C.8 EC2-1-2: 5.3.3(1), Annex C & NA For w = 1.0, e = 0.25b, R120, l = 30 EC2-1-2: and interpolating between n = 0.3 and n = 0.5, b min = 350 mm, a min = 48 mm. = Axis distance = 43 mm + f/2 is OK = 4 no. H32 + 4 no. H25 with 35 mm cover to 8 mm links (a = 55 mm min.) OK Annex C(2) 5.4.3 Structural design: check slenderness about z axis Effective length, l 0 , about z axis: l 0z = 0.5l [1 + k 1 /(0.45 + k 1 ) ] 0.5 [1 + k 2 /(0.45 + k 2 ) ] 0.5 Exp. (5.15) where PD 6687: 2.10 l = clear height between restraints = 4000 – 300/2 = 3850 mm k 1 , k 2 = relative flexibilities of rotational restraints at ends 1 and 2 respectively k 1 = [EI col /l col ]/[2EI beam1 /l beam1 + 2EI beam2 /l beam2 ] ≥ 0.1 Cl. 5.8.3.2(3) where Treating beams as rectangular and cancelling E throughout: I col /l col = 3504/(12 × 3850) = 3.25 × 10 5 I beam1 /l beam1 = 8500 × 300 3 /12 × 6000 = 31.8 × 10 5 I beam2 /l beam2 = 0 PD 6687 k 1 = 3.25/(2 × 31.8) = 0.051 ≥ 0.1 k 1 = 0.1 k 2 = by inspection (pinned end assumed) = ∞ This publication is available for purchase from www.concretecentre.com 160 = l 0z = 0.5 × 3850 × [1 + 0.1/(0.45 + 0.1) ] 0.5 [1 + ∞/(0.45 + ∞)] 0.5 = 0.5 × 3850 × 1.087 × 1.41 = 0.77 × 3850 = 2965 mm Slenderness ratio, l z : Cl. 5.8.3.2(1) l z = l 0z /i where i = radius of gyration = h/3.46 l z = 3.46l 0z /h = 3.46 × 2965/350 = 29.3 Limiting slenderness ratio, l lim : Cl. 5.8.3.1(1) l lim,z = 20 ABC/n 0.5 where A = 0.7 B = 1.1 ‡ C = 1.7 – r m where r m = M 01 /M 02 say M 01 = 0 (pinned end) =r m = 0 C = 1.7 – 0 = 1.7 n = relative normal force = N Ed /A c f cd = 1824.1 × 10 3 /(350 2 × 0.85 × 30/1.5) = 0.88 = l lim,z = 20 × 0.7 × 1.1 × 1.7/0.88 0.5 = 27.9 = As l z > l lim,z column is slender about z axis. Exp. (5.13N) 5.4.4 Check slenderness on y axis Effective length, l 0 , about z axis: l 0y = 0.5l y [1 + k 1 /(0.45 + k 1 ) ] 0.5 [1 + k 2 /(0.45 + k 2 ) ] 0.5 Exp. (5.15) where l y = clear height between restraints = 4000 + 300/2 – 750 = 3400 mm k 1 = relative column flexibility at end 1 = ( I col /I col )/[S2(I beam /I beam )] where I col /I col = 350 4 /12 × 3400 = 3.68 × 10 5 ‡ On first pass the default value for B is used. It should be noted that in the final design w = A s f yd /A c f cd = 6432 × (500/1.15) / (350 2 × 30 × 0.85/1.5) = 2796/2082 = 1.34. So B = (1 + 2 w) 0.5 = (1 + 1.34) 0.5 = 1.92 and the column would not have been deemed ‘slender’. B = 1.1 relates approximately to a column with f ck = 30 MPa and r = 0.4%. * PD 6687 states that to allow for cracking, the contribution of each beam should be taken as 2EI/l beam Cl. 5.8.3.1(1), & NA, EC2-1-2: 5.3.3(2) PD 6687* This publication is available for purchase from www.concretecentre.com 161 S4 Sme|| per|meter co|umn Treating beams as rectangular I beamAB /l beamAB = 350 × 750 3 /[12 × (9000 – 350)] = 14.2 × 10 5 I beamBC /l beamBC = 350 × 750 3 /[12 × (8000 – 350)] = 16.1 × 10 5 k 1 = 3.68/(2 × (16.1 + 14.2) = 0.060 ≥ 0.1 k 1 = 0.1 k 2 = ∞ (pinned end assumed) =l 0y = 0.5 × 3400 [1 + 0.1/(0.45 + 0.1) ] 0.5 [1 +Ƹ/(0.45 + ∞) ] 0.5 = 0.5 × 3400 × 1.087 × 1.41 = 0.77 × 3400 = 2620 mm Exp. (5.15) Slenderness ratio, l y : l y = 3.46l 0y /h = 3.46 × 2620/350 = 25.9 Limiting slenderness ratio, l lim : l lim,y = l lim,z = 27.9 As l y < l lim,y , column not slender in y axis. 5.4.5 Design moments: M Edz about z axis M Edz = max[M 02 ; M 0Ed + M 2 ; M 01 + 0.5M 2 ] Cl. 5.8.8.2 where M 02 = M z + e i N Ed ≥ e 0 N Ed Cl. 5.8.8.2(1), where M z = 146.1 kNm from analysis e i N Ed = effect of imperfections where e i = l 0 /400 6.1.4 Cl. 5.2.7 e 0 = 20 mm = M 02 = 146.1 + (2965/400) × 1824.1 ≥ 0.02 × 1824.1 = 146.1 + 13.4 > 36.5 = 159.5 kNm M 0Ed = equivalent 1st order moment at about z axis at about mid-height may be taken as M 0ez where M 0ez = (0.6M 02 + 0.4M 01 ) ≥ 0.4M 02 = 0.6 × 159.5 + 0.4 × 0 ≥ 0.4 × 159.5 = 95.7 kNm Cl. 5.8.8.2(2) M 2 = nominal 2nd order moment = N ld e 2 Cl. 5.8.8.2(3) where e 2 = (1/r) l 0 2 /10 Cl. 5.8.8. 3 where 1/r = curvature = K v K h [f yd /(E s × 0.45d)] Exp. (5.34) where K v = a correction factor for axial load = (n u – n)/(n u – n bal ) This publication is available for purchase from www.concretecentre.com 162 where n u = 1 + w where w = mechanical ratio = A s f yd /A c f d = 1.08 as before n u = 2.08 n = N Ed /A c f cd = 1824.1/2082 = 0.88 n bal = the value of n at maximum moment resistance = 0.40 (default) K v = (2.08 – 0.88)/(2.08 – 0.40) = 1.20/1.68 = 0.71 K h = a correction factor for creep = 1 + bh ef where b = 0.35 + (f ck /200) – (l/150) = 0.35 + 30/200 – 29.3/150 = 0.35 + 0.15 – 0.195 = 0.305 h ef = effective creep coefficient ‡ Cl. 5.8.4(2) = h (∞,t0) M 0,Eqp /M 0Ed where h (∞,t0) = final creep coefficient Cl. 3.1.4(2) = from Figure 3.1 for inside conditions h = 350 mm, C30/37, t 0 = 15 Fig. 3.1a ƺ 2.4 M 0,Eqp = 1st order moment due to quasi- permanent loads ≈ G k + h 2 Q k × M z + e i N Ed jg G G k + h 0 g Q Q k = 63.3 + 0.8 × 46.0 × M z + e i N Ed 1.35 × 63.3 + 1.5 × 46.0 = 100.1 × 146.1 + 13.4 154.5 = 108.1 kNm M 0Ed = M 02 = 159.5 kNm ‡ With reference to Exp. (5.13N), h ef may be taken as equal to 2.0. However, for the purpose of illustration the full derivation is shown here. Exp. (5.1.3N) This publication is available for purchase from www.concretecentre.com 163 S4 Sme|| per|meter co|umn K h = 1 + 0.305 × 2.4 × 108.1/159.5 = 1.50 f yd = 500/1.15 = 434.8 MPa E s = 200000 MPa Cl. 3.2.7(3) d = effective depth = 350 – 35 – 8 – 16 = 291 mm 1/r = 0.71 × 1.50 × 434.8/(200000 × 0.45 × 291) = 0.0000177 l 0 = 2965 mm as before e 2 = (1/r) l 0 2 /10 = 0.0000177 × 2965 2 /10 = 15.6 mm =M 2 = N Ed e 2 = 1824.1 × 10 3 × 15.6 = 28.4 kNm M 01 = 0 = M Edz = max[M 02z ; M 0Edz + M 2 ; M 01 + 0.5M 2 ] = max[159.5; 95.7 + 28.4; 0 + 28.4/2] = 159.5 kNm a) 1st order moments from analysis M z = 146.1 kNm M 2 = 28.4 kNm M 2 /2 = 14.2 kNm M OEdz = 95.8 kNm M z = 0 M Edz = 159.5 kNm M z = 0 e i N Ed = 13.4 kNm b) Including 2nd order moments: M Edz = max [M 02 , M OEd + M 2 , M 01 = 0.5M 2 ] c) Design moments: M Edz about z axis + = Figure 5.13 Design moments M Edz 5.4.6 Design moments: M Edy about y axis M Edy = max[ M 02y ; M 0Edy + M 2 ; M 01 + 0.5M 2 ] where M 02y = M y + e i N Ed ≥ e 0 N Ed = 114.5 + 13.4 § ≥ 36.7 kNm = 127.9 kNm M 0Edy = (0.6M 02y + 0.4 M 01y ) ≥ 0.4M 02y = 0.6 × 114.5 + 0.4 × 0 § Imperfections need to be taken into account in one direction only. Cl. 5.8.9(2) This publication is available for purchase from www.concretecentre.com 164 = 68.7 kNm M 2 = 0 (as column is not slender not slender about y axis). = M Edy = 127.9 kNm 5.4.7 Design in each direction using charts In z direction: N Ed /bhf ck = 1824.1 × 10 3 /(350 2 × 30) = 0.50 M Ed /bh 2 f ck = 159.5 × 10 6 /(350 3 × 30) = 0.124 Assuming 8 bar arrangement, centroid of bars in half section: d 2 ≥ 35 + 8 + 16 + (350/2 – 35 –8 – 16) × 1/4 Fig. C4e) ƽ 59 + 29 = 88 mm d 2 /h = 0.25 From Figure C4e) A s f yk /bhf ck = 0.50 A s = 0.50 × 350 2 × 30/500 = 3675 mm 2 = 4 no. H32 + 4 no. T25 (5180 mm 2 ) OK. Fig. C4e) In y direction: M Ed /bh 2 f ck = 127.9 × 10 6 /(350 3 × 30) = 0.10 N Ed /bhf ck = 0.50 From Figure C4e) A s f yk /bhf ck = 0.34 A s = 0.34 × 3502 × 30/500 = 2499 mm 2 = 4 no. H32 + 4 no. T25 (5180 mm 2 ) OK. 5.4.8 Check biaxial bending l y ≈ l z = OK. Exp. (5.38a) e z = M Edy /N Ed e y = M Edz /N Ed e y /h eq = M Edz = 159.5 = 1.25 e z /b eq M Edy 127.9 Exp. (5.38b) = need to check biaxial bending (M Edz /M Rdz ) a + (M Edy /M Rdy ) a ≤ 1.0 where Exp. (5.39) M Rdz = M Rdy = moment resistance= Using Figure C4e) Fig. C4e) A s f yk /bhf ck = 5180 × 500/(350 2 × 30) = 0.70 for N Ed /bhf ck = 0.50 M Ed /bh 2 f ck = 0.160 = M Rd = 0.160 × 350 3 × 30 This publication is available for purchase from www.concretecentre.com 165 S4 Sme|| per|meter co|umn = 205.8 kNm a depends on N Ed /N Rd where N Rd = A c f cd + A s f yd = 350 2 × 0.85 × 30/1.5 + 5180 × 500/1.15 = 2082.5 + 2252.2 = 4332.7 kN N Ed /N Rd = 1824.1/4332.7 = 0.42 = a = 1.27 (159.5/205.8) 1.27 + (114.5/205.8) 1.27 = 0.72 + 0.47 = 1.19 = No good = Try 8 no. T32 (6432 mm 2 ) Cl. 5.8.9(4) For A s f yk /bhf ck = 6432 × 500/(350 2 × 30) = 0.88 for N Ed /bhf ck = 0.50 M Ed /bh 2 f ck = 0.187 Fig. C4e) = M Rd = 240.5 kNm Check biaxial bending (159.5/245.7) 1.27 + (114.5/245.7) 1.27 = 0.59 + 0.39 = 0.98 OK 5.4.9 Check maximum area of reinforcement A s /bd = 6432/350 2 = 5.2% > 4% Cl. 9.5.2(3) & NA However, if laps can be avoided in this single lift column then the integrity of the concrete is unlikely to be affected and 5.2% is considered OK. OK PD 6687: 2.19 5.4.10 Design of links Diameter min. = 32/4 = 8 mm Cl. 9.5.3 & NA Spacing max. = 0.6 × 350 = 210 mm Cl. 9.5.3(3), 9.5.3(4) = Use H8 @ 200 mm cc 5.4.11 Design summary 8 H32 H8 links @ 200 cc 35 mm cover to link No laps in column section Note The beam should be checked for torsion. Figure 5.14 Design summary: small perimeter column This publication is available for purchase from www.concretecentre.com 166 Walls General ve||s ere deuned es |e|ng vert|ce| e|ements whose |engths ere ¦our t|mes greeter then the|r th|c|nesses ¹he|r des|gn does not d|¦¦er s|gn|ucent|y ¦rom the des|gn o¦ co|umns |n thet ex|e| |oeds end moments e|out eech ex|s ere essessed end des|gned ¦or ¹he ce|cu|et|ons |n th|s sect|on |||ustrete the des|gn o¦ e s|ng|e sheer we|| Cenere||y, the method o¦ des|gn|ng we||s |s es ¦o||ows ln prect|ce, severe| o¦ these steps mey |e com||ned Leterm|ne des|gn ||¦e N EC0 & NA Table NA 2.1 Assess ect|ons on the we|| N EC1 (10 parts) & UK NAs Leterm|ne wh|ch com||net|ons o¦ ect|ons epp|y N EC0 & NA: Tables NA A1.1 & NA: A1.2(B) Assess dure||||ty requ|rements end determ|ne concrete N strength BS 8500–1 Chec| cover requ|rements ¦or eppropr|ete ¦|re N res|stence per|od Approved Document B EC2–1–2 Leterm|ne cover ¦or ¦|re, dure||||ty end |ond N Cl. 4.4.1 Ane|yse structure ¦or cr|t|ce| com||net|on moments N end ex|e| ¦orces Section 5 Chec| s|enderness end determ|ne des|gn moments N Section 5.8 Leterm|ne eree o¦ re|n¦orcement requ|red N Section 6.1 Chec| spec|ng o¦ |ers N Sections 8 & 9 6 6.0 This publication is available for purchase from www.concretecentre.com ve||s 167 6.1 Shear wall lxemp|e 6! shows the des|gn o¦ e s|mp|e ||neer sheer we|| es typ|ce||y used |n med|um- r|se |u||d|ngs S|m||er pr|nc|p|es mey |e epp||ed to we||s thet ere sheped es C, l, ¹, ? end recteng|es |n-p|en, |ut |ssues o¦ ||m|t|ng ¦enge d|mens|ons end sheer et corners need to |e eddressed ¹he exemp|e shows on|y 0lS des|gn es, epert ¦rom m|n|mum erees o¦ stee| to contro| crec||ng, SlS |ssues ere genere||y non-cr|t|ce| |n med|um-r|se structures lor sheer we||s |n h|gh-r|se structures, re¦erence shou|d |e mede to spec|e||st ||tereture |?9| ¹he exemp|e |s |ntended to show how e sheer we|| prov|d|ng pert o¦ the |etere| ste||||ty |n one d|rect|on |n e med|um r|se structure m|ght |e des|gned |y hend Ax|e| |oeds end urst order moments ere determ|ned ¹he des|gn cons|ders s|enderness |n order to determ|ne des|gn moments, / ld , |n the p|ene perpend|cu|er to the we|| ¹he e¦¦ects o¦ e||ow|ng ¦or |mper¦ect|ons ere e|so |||ustreted lro|ect dete||s Ce|cu|eted |y chg ¦o| no CCIP – 041 Chec|ed |y web Sheet no 1 C||ent TCC Lete Oct 09 Shear wall Wall A is 200 mm thick and, in addition to providing vertical support to 200 mm flat slabs at roof level and floors 1 to 3, it helps to provide lateral stability to the four-storey office block. Assuming the stair itself provides no lateral stability, the wall is to be designed for the critical section at ground and first floor level using BS EN 1990 Exp. (6.10). The concrete is C30/37. The wall is supported on pad foundations and the ground floor is ground bearing. Wall A 1300 4400 1500 7200 300 3600 6000 6000 6000 6000 2500 300 300 4800 4800 4800 300 30700 N X X Figure 6.1 Typical floor plan Figure 6 1 Typical floor plan 900 Roof 3rd 2nd 1st Gnd 4@3300 600 900 Figure 6.2 Section X–X This publication is available for purchase from www.concretecentre.com 168 6.1.1 Actions Permanent actions Variable actions g k q k kN/m 2 Roof Paving 40 mm Waterproofing Insulation Suspended ceiling Services Self-weight 200 mm slab Imposed load 1.00 0.50 0.10 0.15 0.30 5.00 7.05 0.60 Section 2.8 Section 2.4.2 Floor slabs Carpet Raised floor Suspended ceiling Services Self-weight 200 mm slab Imposed load 0.03 0.30 0.15 0.30 5.00 5.78 2.50 Section 2.8 Section 2.4.2 Ground floor slab (ground bearing) Carpet Raised floor Services Self-weight 200 mm slab Imposed load 0.03 0.30 0.15 5.00 5.48 3.00 Section 2.4.2 Stairs 150 waist @ 30 Treads 0.15 × 0.25 × 25 × 4/2 = Screed 0.05 × 22 = Plaster Tiles and bedding Imposed load 4.40 1.88 1.10 0.21 1.00 8.59 2.50 Section 2.8 Section 2.4.2 Cavity wall 102 mm brickwork 50 mm insulation 100 mm blockwork Plaster 2.37 0.02 1.40 0.21 4.00 Section 2.8 RC wall 200 mm wall Plaster both sides 5.00 0.42 5.42 Section 2.8 Wind w k = 1.10 EC1-1-4 & NA This publication is available for purchase from www.concretecentre.com 169 6! Sheer ve|| 6.1.2 Load take-down Consider whole wall. G k Q k Item Calculation From item Cum. total From item Cum. total Roof (6.0/2 + 2.5/2) × (4.4 + 1.5/2) × (7.05 + 0.6) = 154.3 13.1 Roof (6.0/2) × (1.3/2) × (7.05 + 0.6) = 13.7 1.2 Wall 3.3 × 4.4 × 5.42 = 78.7 246.7 14.3 At above 3rd floor 246.7 14.3 3rd floor (6.0/2) × (1.3/2 + 4.4 + 1.5/2) × (5.78 + 2.5) = 100.6 43.5 Landing (2.5/2 × 1.5/2) × (5.78 + 2.5) 11.6 5.0 Wall a. b. 78.7 Stair say 1.1 × 4.4 (8.59 + 2.5) 41.6 12.1 232.5 60.6 At above 2nd floor 479.2 74.9 2nd floor, landing, wall and stair a. b. 232.5 60.6 At above 1st floor 711.7 135.5 1st floor, landing, wall and stair a. b. 232.5 60.6 At above ground floor 944.2 196.1 Ground floor assume 1 m all round = 2 × (1.3/2 + 4.40 + 1.5/2) × (5.48 + 3.0) = 63.6 34.8 250 mm wall to foundation 4.4 × 0.2 × 0.6 × 25 = 13.2 76.8 At above foundation 1021.0 230.9 6.1.3 Design actions due to vertical load at ground–1st G k = 944.2 G k /m = 944.2/4.4 = 214.6 kN/m Q k = a n × 196.1 where a n = 1.1 – n/10 where n = no. of storeys qualifying for reduction ‡ = 3 = 1.1 – 3/10 = 0.8 = Q k = 0.8 × 196.1 = 156.9 kN Q k /m = 156.9/4.4 = 35.7 kN/m EC1-1-1: 6.3.1.2(11) & NA ‡ Includes storeys supporting Categories A (residential and domestic), B (office), C (areas of congregation) and D (shopping), but excludes E (storage and industrial), F (traffic), G (traffic) and H (roofs). This publication is available for purchase from www.concretecentre.com 170 6.1.4 Vertical loads from wind action: moments in plane Consider wind loads, N–S Wall A Lift shaft 200 thick walls 300 3600 6000 6000 6000 6000 2500 300 2400 o/a 2400 4400 30700 N w k = 1.10 kN/m 2 Figure 6.3 Lateral stability against wind loads N–S Check relative stiffness of lift shaft and wall A to determine share of load on wall A. Lift shaft: I LS = 2.4 4 /12 – 2.0 4 /12 – 0.2 × 1.6 3 /12 = 1.36 m 4 Wall A: I WallA = 0.2 × 4.4 3 /12 = 1.41 m 4 where I = inertia = Wall A takes 1.41/(1.41 + 1.36) = 51% of wind load. Check shear centre to resolve the effects of torsion. Determine centre of gravity, CoG L of the lift shaft. Area, A Lever arm, x Ax 2.4 × 2.4 = 5.76 1.2 6.912 –2.0 × 2.0 = –4.00 1.2 –4.800 –1.6 × 0.2 = –0.32 2.3 –0.732 1.44 1.38 2400 CoG L x 2 4 0 0 1 6 0 0 2 0 0 2 0 0 Figure 6.4 Lift shaft This publication is available for purchase from www.concretecentre.com 171 6! Sheer we|| x = Ax/A = 1.38/1.44 = 0.956 m i.e. from face of lift shaft to CoG of shaft = 2.40 – 0.956 = 1.444 m Shear centre, C w of walls, from centreline of wall A = I LS × (1.44 + 24.00 + 0.05 ‡ ) = 1.36 × 25.49 = 12.56 m from wall A I LS + I WallA 1.36 + 1.41 or = 12.56 + 2.80 – 0.05 = 15.31 from east end of building. 1.45 24.00 2.80 C Wall A ‡ 15.31 15.35 W k CoG L * C w * L Wall A Figure 6.5 Shear centre, C w CC and centre of action, w W k WW Centre of action (30.7/2 = 15.35 m from end of building) and shear centre (almost) coincide. = there is no torsion to resolve in the stability system for wind in a N–S direction. # 900 Roof 3rd 2nd 1st Gnd 3300 3300 3300 3300 900 600 1 4 1 0 0 w k = 17.2 kN/m 300 4400 300 Figure 6.6 Wall A – wind loads N–S = Wall A takes 51% of wind load, so characteristic wind load on wall A, w k, wall A = 51% × w k × L x = 51% × 1.1 × 30.7 = 17.2 kN/m ‡ Assuming centreline of wall A is 50 mm to right hand side of grid. # Had there been significant torsion this would have been resolved into +/– forces in a couple based on the shear walls. This publication is available for purchase from www.concretecentre.com 172 = at just above ground floor, characteristic in-plane moment in wall A, M k , due in this case to wind = 17.2 × 14.1 2 /2 = 1709.8 kNm Resolving into couple using 1 m either end of wall ‡ , characteristic wind load in each end, W k = 1709.8/3.4 = ±502.9 kN 6.1.5 Effects of global imperfections in plane of wall A H IG H I1 H I2 H I3 H IR y I Roof 3rd 2nd 1st Gnd Figure 6.7 Global imperfections Global imperfections can be represented by forces H i at floor level where H i = y i (N b – N a ) Exp. (5.4) where y i = (1/200) a h a m Cl. 5.2(1), 5.2(5), 5.2(8) & NA where a h = 0.67 ≤ 2/l 0.5 ≤ 1.0 = 0.67 ≤ 2/14.7 0.5 ≤ 1.0 = 0.67 ≤ 0.52 ≤ 1.0 = 0.67 a m = [0.5(1 + 1/m)] 0.5 where m = no. of members contributing to the total effect = 25 vertical elements on 4 floors = 100 ‡ For medium-rise shear walls there are a number of methods of design. Cl. 9.6.1 suggests strut-and-tie (see Volume 2 of these worked examples [30] ). Another method [26] is to determine elastic tensile and compression stresses from N Ed /bL +/– 6M Ed /bL 2 and determine reinforcement requirements based on those maxima. The method used here assumes a couple, consisting of 1.0 m of wall either end of the wall. The reinforcement in tension is assumed to act at the centre of one end and the concrete in compression (with a rectangular stress distribution) acts at the centre of the other end. The forces generated by the couple add or subtract from the axial load in the 1 m ends of the walls. The method is useful for typical straight shear walls of say 2.5 to 5.0 m in length. Vol. 2 This publication is available for purchase from www.concretecentre.com 173 6! Sheer we|| = a m = 0.71 = y i = 0.67 × 0.71/200 = 0.0024 N b , N a = axial forces in members below and above (N b – N a ) = axial load from each level At roof level Area = 30.4 × 14.5 – 1.3 × 2.5 – 3.6 × 4.8 = 420.3 m 3 Perimeter = 2 × (30.4 + 14.5) = 89.8 m (N a – N b ) = axial load from roof level = 420.3 × (7.05 + 0.6) + 89.8 × 0.9 × 4.0 = 3286.4 + 252.2 kN At 3rd floor (N a – N b ) = 420.3 × (5.78 + 2.5) + 89.8 × 3.3 × 4.0 = 3615.7 + 1050.8 kN At 2nd floor (N a – N b ) = 3615.7 + 1050.8 kN At 1st floor (N a – N b ) = 3615.7 + 1050.8 kN H iR = 0.0024 × (3286.4 + 252.2) = 7.9 + 0.6 = 8.5 kN H i3 = H i2 = H i1 = 0.0024 × (3615.7 + 1050.8) = 8.7 + 2.5 = 11.2 kN Characteristic design moment at ground floor, M k = 8.5 × 13.2 + 11.2 × (9.90 + 6.60 + 3.30) = 112.2 + 221.8 = 334.0 kNm As before, wall A resists 51% of this moment. Resolving into couple using 1 m either end of wall, = G kH § = 0.51 × 334.0/3.4 = ± 50.1 kN i.e. G kH = ±50.1 kN/m 6.1.6 Check for global second order effects To check whether the building might act as a sway frame check Cl. 5.8.3.3(1) F V,Ed ≤ k l n s = SE cd I c n s + 1.6 L 2 Exp. (5.18) where F V,Ed = Total vertical load (on braced and bracing members) where Floor area = (30.7 – 2 × 0.15) × 14.4 – (2 × 0.15) – 3.6 × 4.8 – 1.3 × 2.5 = 428.6 – 20.5 = 408.1 § As H i derives mainly from permanent actions its resulting effects are considered as being a permanent action too. This publication is available for purchase from www.concretecentre.com 174 Loads G k Q k from roof: 408 (7.05 + 0.6) = 3–1 floors: 3 × 408 (5.78 + 2.5) = Allow cavity wall at 1st floor and above 2876 7075 245 3060 (3 × 3.30 + 0.9) × 2 × (30.4 + 14.1) × 4.0 = Imposed load reduction 20% (see 6.2.3) 3845 13705 13705 3305 661 2644 =F V,Ed ≈ 13705 × 1.35 + 1.5 × 2644 = 22468 kN k l = 0.31 n s = number of storeys = 4 (including roof) E cd = E cm /g CE = 33/1.2 = 27.5 GPa Cl. 5.8.2(6) & NA Table 3.1, 5.8.6(3) & NA I c = Inertia of bracing members in N–S direction I c = 1.36 + 1.41 = 2.77 m 4 (See Section 6.1.4) in E–W direction I LS , with reference to Figure 6.4 h × d Area, A × Ax Ax 2 I 2.4 × 2.4 = 5.76 1.2 6.912 8.294 2.765 –2.0 × 2.0 = –4.00 1.2 –4.800 –5.760 –1.333 –1.6 × 0.2 = –0.32 2.3 –0.732 –1.683 –0.001 1.44 1.38 0.851 1.431 as before (6.1.4), x = 1.38/1.44 = 0.956 m I LS = I NU = Ax 2 + I – Ax 2 = 0.851 + 1.431 – 1.44 × 0.956 = 0.965 m 4 L = total height of building above level of moment restraint = 14.7 (see Figure 6.6) Check k l n s × SE cd I c on weak E–W axis: n s + 1.6 L 2 = 0.31 × [4/(4 + 1.6)] × 27500 × 10 3 × (0.965/14.7 2 ) = 27200 kN i.e. > F V,Ed = no need to design for 2nd order effects. This publication is available for purchase from www.concretecentre.com 175 6! Sheer we|| 6.1.7 Design moments – perpendicular to plane of wall 1 3 5 0 6 5 0 4 4 0 0 7 5 0 C L C L C L 150 B A A B up 200 1st Gnd Figure 6.8 Plan of wall A and location of sections A–A and B–B Figure 6.9 Section A–A Section A–A @ 1st floor The slab frames into the wall. For the purposes of assessing fixed end moments, the width of slab contributing to the moments in the wall is assumed to be the length of the wall plus distances half way to adjacent supports either end. Therefore, consider the fixed end moment for 1.50/2 + 4.40 + 1.30/2 = 5.8 m width of adjoining slab framing into the 4.4 m long shear wall (see Figure 6.8). n k s k w k w Figure 6.10 Subframe section A–A @ 1st floor FEM ‡ : assuming imposed load is a leading variable action: = nl 2 /8 = 5.8 (1.35 × 5.78 + 1.5 × 2.5) × 6.0 2 /8 = 5.8 × 11.6 × 6 2 /8 = 302.8 kNm EC0: Exp. (6.10) & NA ‡ FEM: fixed end moment This publication is available for purchase from www.concretecentre.com 176 k w = EI/l = E × 4400 × 200 3 /(12 × 3300) = E × 8.88 × 10 5 k s = EI/2l = E × 5800 × 200 3 /(2 × 12 × 6000) = E × 3.22 × 10 5 M = 302.8 × 8.88/(2 × 8.8 + 3.22) = 302.8 × 0.42 = 121.2 kNm i.e. 121.2/4.40 = 27.5 kNm/m @ ULS Similarly, assuming imposed load is an accompanying action: FEM = 5.8 (1.35 × 5.78 + 0.7 × 1.5 × 2.5) × 6 2 /8 = 5.8 × 10.4 × 6 2 /8 = 271.4 kNm =M = 271.4 × 0.42/4.40 = 25.9 kNm/m @ ULS Section A–A @ ground floor By inspection not critical – nominal moment. Section B–B @ 1st Consider the landing influences half of wall (2.2 m long) and that this section of wall is subject to supporting half the slab considered before at 1st floor level at Section A–A. 1 6 5 0 1st Landing Gnd Figure 6.11 Section B–B FEM = 302.8/2 = 151.4 kNm k w = I/l = 2200 × 200 3 /(12 × 1650) = 8.88 × 10 5 k s = 3.22 × 10 5 /2 = 1.61 × 10 5 M = 151.4 × 8.88/(2 × 8.88 + 1.61) = 151.4 × 0.46 = 69.6 kNm i.e. 63.8/2.2 = 31.6 kNm/m @ ULS This publication is available for purchase from www.concretecentre.com 177 6! Sheer we|| Similarly, assuming imposed load is an accompanying action: FEM = 5.8 (1.35 × 5.78 + 0.7 × 1.5 × 2.5) × 6 2 /8 = 5.8 × 10.4 × 6 2 /8 = 271.4 kNm =M = 271.4 × 0.46/(2 × 2.2) = 28.4 kNm/m @ ULS Section B–B @ landing level and ground floor By inspection not critical 6.1.8 Consider slenderness of wall at ground floor To derive maximum slenderness (at south end of wall), ignore effect of landing. Effective length, l 0 = 0.75 × (3300 – 200) = 2325 Table C16 l = 3.46 × l 0 /h = 3.46 × 2325/200 = 40.2 Cl. 5.8.3.2(1) Limiting slenderness, l lim = 20 ABC/n 0.5 Cl. 5.8.3.1(1), Exp. (5.13N) where A = 0.7 B = 1.1 C = 1.7 – r m where r m = M 01 /M 02 = say = –0.25 C = 1.95 n = N Ed /A c f d where N Ed = 214.6 × 1.25 + 31.2 × 1.5 × 0.7 + 502.9 × 1.5 + 98.2 × 1.5 × 0.7 ‡ = 268.3 + 32.8 + 754.4 + 103.1 = 1158.6 kN A c f d = 200 × 1000 × 0.85 × 30/1.5 = 3400 kN = n = 0.34 = l lim = 20 × 0.7 × 1.1 × 1.95/0.34 0.5 = 51.5 = As l < l lim wall is not slender and = no secondary moments 6.1.9 Summary: design forces on wall, ground–1st floor At ground to 1st consider maxima. Vertical loads G k = 214.6 kN/m Q k = 35.7 kN/m Vertical load due to in-plane bending and wind W k = ±502.9 kN/m Vertical load due to in-plane bending and imperfections G kH = ±50.1 kN/m Maximum moment out of plane, floor imposed load as leading action M = 31.6 kN/m @ ULS Maximum moment out of plane, floor imposed load as accompanying action M = 28.4 kN/m @ ULS ‡ Assuming wind load is lead variable action. This publication is available for purchase from www.concretecentre.com 178 6.1.10 Combinations of actions at ground–1st floor a) At ULS, for maximum axial load, W k is leading variable action N Ed = 1.35G k + 1.5Q k1 + 1.5c 0 Q ki = 1.35 (214.6 + 50.1) + 1.5 × 502.9 + 1.5 × 0.7 × 35.7 = 357.3 + 754.4 + 37.5 = 1149.2 kN/m M Ed = M + e i N Ed ≥ e 0 N Ed Cl. 5.8.8.2(1), 6.1.4 where M = moment from 1st order analysis = 28.4 kNm/m e i = l 0 /400 = 2325/400 = 5.8 mm Cl. 5.2(7), 5.2(9) e 0 = h/30 ≥ 20 mm = 20 mm Cl. 6.1.4 M Ed = 28.4 + 0.0058 × 1149.2.1 ≥ 0.020 × 1149.2 = 28.4 + 6.7 ≥ 23.0 = 35.1 kNm/m b) At ULS, for minimum axial load, W k is leading variable action N Ed = 1.0 × 214.6 – 1.35 × 50.1 – 1.5 × 502.9 + 0 × 35.7 = –607.4 kN/m (tension) M Ed = 28.4 ‡ + 0.0058 × 607.4 ≥ 0.020 × 602.4 = 28.4 + 3.5 ≥ 23.0 = 31.9 kNm/m c) At ULS, for maximum out of plane bending assuming Q k is leading variable action N Ed = 1.35 (214.6 + 50.1) + 1.5 × 35.7 + 1.5 × 0.5 × 502.9 = 357.3 + 53.6 + 377.2 = 788.1 kN/m M Ed = 31.6 + 0.0058 × 788.1 ≥ 0.020 × 788.1 = 31.6 + 4.6 ≥ 15.8 = 36.2 kNm/m or N Ed = 1.0 × 214.6 – 1.35 × 50.1 – 0 × 31.2 – 1.5 × 0.5 × 502.9 = 214.6 – 67.6 – 0 – 377.2 = –230.2 kN/m (tension) M Ed = 31.6 + 0.0058 × 230.2 = 33.0 kNm/m d) Design load cases Consolidate c) into a) and b) to consider two load cases: N Ed = 1149.4 kN/m, M Ed = 36.2 kN/m (out of plane) and N Ed = –607.4 kN/m, M Ed = 36.2 kN/m (out of plane) ‡ Strictly incompatible with Q k = 0. However, allow Q k = 0. This publication is available for purchase from www.concretecentre.com 179 6! Sheer we|| 6.1.11 Design: cover above ground c nom = c min + Dc dev where c min = max[c min,b ; c min,dur ] Exp. (4.1) where c min,b = diameter of bar = 20 mm vertical or 10 mm lacers c min,dur = for XC1 = 15 mm Dc dev = 10 mm =c nom = 15 + 10 = 25 mm to lacers (35 mm to vertical bars) 6.1.12 Fire resistance Assuming 1-hour fire resistance required for, as a worst case, μ fi = 0.7 and fire on both sides. Min. thickness = 140 mm, min. axis distance = 10 mm i.e. not critical EC2-1-2: Table 5.4 6.1.13 Design using charts For compressive load: d 2 /h = (25 + 10 + 16/2)/200 = 0.215 = interpolate between charts C5d) and C5e) for Figs C5d), C5e) N Ed /bhf ck = 1149.4 × 10 3 /(200 × 1000 × 30) = 0.192 M Ed /bh 2 f ck = 36.2 × 10 6 /(200 2 × 1000 × 30) = 0.030 Gives: A s f yk /bhf ck = 0 = minimum area of reinforcement required = 0.002 A c Cl. 9.6.2 & NA = 0.002 × 200 × 1000 = 400 mm 2 /m = 200 mm 2 /m each face max. 400 mm cc, min. 12 mm diameter Try T12 @ 400 Cl. 9.6.2(3); SMDSC A s /2 Moment 0 0 0 0 Tension Combined + e e e s s st2 s st1 › › + = = Combined A s /2 Figure 6.12 Stresses and strains in wall subject to tension and out of plane moment p This publication is available for purchase from www.concretecentre.com 180 For tensile load and moment: Working from first principles, referring to Figure 6.12 and ignoring contribution from concrete in tension, N Ed = (s st1 + s st2 ) × A s /2 and M Ed = (s st1 – s st2 ) × A s /2 × (d – d 2 ) so s st1 + s st2 = 2N Ed /A s and s st1 – s st2 = 2M Ed /[(d – d 2 )A s ] = 2s st1 = 2N Ed /A s + 2M Ed /[(d – d 2 )A s ] = A s = (N Ed /s st1 ) + M Ed /(d – d 2 )s st1 s st1 = f yk /g S = 500/1.15 = 434.8 = A s = 607.4 × 10 3 /434.8 + 36.2 × 10 6 /[(157 – 43) × 434.8] = 1397 + 730 = 2127 mm 2 s st2 = 2N Ed /A s – s st1 = 571.7 – 434.8 = 136 MPa By inspection all concrete is in tension zone and may be ignored. Use 6 no. H16 @ 200 cc both sides for at least 1 m each end of wall (2412 mm 2 ). 6.1.14 Horizontal reinforcement A s, hmin = 0.001A s or 25% A s vert = 200 mm 2 or 0.25 × 2036 = 509 mm 2 /m = requires 254 mm 2 /m each side Cl. 9.6.3(1) & NA Spacing ≤ 400 mm Cl. 9.6.3(2) Links not required. Use H10 @ 300 (262 mm 2 /m) both sides. Cl. 9.6.4(1) 6.1.15 Check for tension at top of foundation Permanent and variable: G k = 1021.0/4.4 = 232.0 kN/m Q k = 230.9/4.4 = 52.5 kN/m Section 6.1.2 Wind: M k = 17.2 × 14.1 × [14.1/2 + 0.6] = 1855.3 kN/m Resolved into couple 1 m either end of wall W kw = 1855.3/3.4 = ±545.7 kN/m Section 6.1.4 Global imperfections: M k = 8.5 × 13.8 + 11.2 × (10.5 + 7.2 + 3.9 + 0.6) = 365.9 kNm G kH = 365.9 × 0.51/3.4 = 54.9 kN/m Section 6.1.5 This publication is available for purchase from www.concretecentre.com 181 6! Sheer we|| At ULS for maximum axial tension W k is lead imposed load: N Ed = 1.0 × 232.0 – 1.35 × 54.9 – 1.5 × 545.7 + 0 × 52.5 = –660.7 kN/m M Ed = nominal = e2N Ed = 0.02 × 660.7 Cl. 6.1.4 = 13.2 kNm/m As before A s = N Ed + M Ed f yk /g M (d – d 2 )f yk /g M = 660.7 × 10 3 /434.8 + 13.2 × 10 6 /[(157 – 43) × 434.8] = 1520 + 266 = 1786 mm 2 i.e. not critical = Use 6 no. H16 @ 200 cc b.s. for at least 1 m either end of wall (2412 mm 2 ). 6.1.16 Check for axial compression at top of foundation At ULS for maximum axial compression W k is lead imposed load: N Ed = 1.35 × 232.0 + 1.35 × 54.9 + 1.5 × 545.7 + 0.7 × 1.5 × 52.5 = 1261.0 kN/m M Ed = nominal = e2N Ed = 0.02 × 1261.0 Cl. 6.1.4 = 25.2 kNm/m By inspection not critical (minimum reinforcement required). Section 6.1.13 = tension critical as above. 6.1.17 Design: cover below ground c nom = c min + Dc dev where c min = max[c min,b ; c min,dur ] Exp. (4.1) where c min,b = diameter of bar = 16 mm vertical or 10 mm lacers c min,dur = for assumed Aggressive Chemical Environment for Concrete (ACEC) class AC1 ground conditions = 25 mm BS 8500-1 Annex A [14] , How to: Building structures [8] Dc dev = 10 mm c nom = 25 + 10 = 35 mm to lacers (45 mm to vertical bars) In order to align vertical bars from foundation into Gnd–1st floor lift as starter bars, locally increase thickness of wall to say 250 mm thick with c nom = 50 mm This publication is available for purchase from www.concretecentre.com 182 6.1.18 Check stability Assume base extends 0.3 m beyond either end of wall A, i.e. is 5.0 m long and is 1.2 m wide by 0.9 m deep. Overturning moments Wind (see Figure 6.6) EC0: Table A1.2(A) & NA Fig. 6.6 M k = 17.2 × 14.1 × [14.1/2 + 1.5] = 2073.5 kNm Global imperfections (see Section 6.1.5) Fig. 6.7 M k = 0.51 x [8.5 × 14.7 + 11.2 × (11.4 + 8.1 + 4.8 + 1.5)] = 0.51 x [125.0 + 11.2 × 25.8] = 0.51 x 414.0 = 211 kNm Restoring moment M k = (1021.0 + 5.0 x 1.2 x 0.9 x 25 + 0 x 230.9 ) x (0.3 + 2.2) = 2890 kNm At ULS of EQU, Overturning moment = fn(g Q,1 Q k1 + g G,sup G k ) = 1.5 x 2073.5 + 1.1 x 211.0 = 3342.4 kNm EC0: Table A1.2(A) & NA Restoring moment = fn(g G,inf G k ) = 0.9 x 2890 = 2601 kNm i.e. > 1818.4 kNm = no good EC0: Table A1.2(A) & NA Try 1.05 m outstand Restoring moment M k = 2890 (1.05 + 2.2) / (0.3 + 2.2) = 3757.0 kNm At ULS, restoring moment = 0.9 x 357.0 = 3381.3 kNm =OK. Use 1.05 m outstand to wall. 6.1.19 Design summary H12 @ 400 b.s 12 12 1st 1050 1050 25 mm cover to 200 mm wall above ground floor 24H16 @ 200 (12NF, 12FF) Gnd 50 mm cover to 250 wall below ground floor Lacers H10 @ 300 m outside This publication is available for purchase from www.concretecentre.com 183 / ke¦erences end ¦urther reed|ng 7 References and further reading References 1 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99?÷!÷!, lurocode ? ÷ lert !÷! /e·/qn o| conc·e|e ·|·ac|a·e· - Oene·o/ ·a/e· onJ ·a/e· |o· /a//J/nq· bSl, ?004 1a Net|one| Annex to lurocode ? ÷ lert !÷! bSl, ?00S 2 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99?÷!÷?, lurocode ? ÷ lert !÷? /e·/qn o| conc·e|e ·|·ac|a·e· - Oene·o/ ·a/e· - ´|·ac|a·o/ |/·e Je·/qn bSl, ?004 2a Net|one| Annex to lurocode ? ÷ lert !÷? bSl, ?00S 3 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99?÷?, lurocode ? ÷ lert ? /e·/qn o| conc·e|e ·|·ac|a·e· - S·/Jqe· bSl, ?00S 3a Net|one| Annex to lurocode ? ÷ lert ? bSl, ?00/ 4 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99?÷3, lurocode ? ÷ lert 3 /e·/qn o| conc·e|e ·|·ac|a·e· - //¡a/J-·e|o/n/nq onJ con|o/nmen| ·|·ac|a·e· bSl, ?006 4a Net|one| Annex to lurocode ? ÷ lert 3 bSl, ?00/ 5 k S NAkA¥ANAN 8 C l COOLClllL ¹he Concrete Centre Conc/·e /a·ocoJe ? CCll-00S ¹CC, ?006 6 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON lL 66S/ Soc/q·oanJ oooe· |o |/e c/ /o|/ono/ /nne\e· S´ // '99?-' bSl, ?006 7 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS S!!0 ´|·ac|a·o/ a·e o| conc·e|e - /o·| ' CoJe o| o·oc|/ce |o· Je·/qn onJ con·|·ac|/on bSl,!99/ 8 bkOOllk, O et e| /ou |o Je·/qn conc·e|e ·|·ac|a·e· a·/nq /a·ocoJe ? CCll-006 ¹he Concrete Centre, ?006 0pdeted ¦or down|oed on|y ?009 9 ¹ll lNS¹l¹0¹lON Ol S¹k0C¹0kAl lNClNllkS/¹ll CONCkl¹l SOCll¹¥/L¹l ´|onJo·J me|/oJ o| Je|o///nq ·|·ac|a·o/ conc·e|e ¹h|rd ld|t|on lStructl, ?006 10 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !990, lurocode So·/· o| ·|·ac|a·o/ Je·/qn bSl, ?00? 10a Net|one| Annex to lurocode bSl, ?004 11 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99!, lurocode ! /c|/on· on ·|·ac|a·e· (!0 perts) bSl, ?00?÷?006 11a Net|one| Annexes to lurocode ! bSl, ?00S÷?00S end |n preperet|on 12 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99/÷!, lurocode / Oeo|ec/n/co/ Je·/qn Oene·o/ ·a/e· bSl, ?004 12a Net|one| Annex to lurocode / bS lN !99/÷! bSl, ?00/ 13 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99S÷!, lurocode S /e·/qn o| ·|·ac|a·e· |o· eo·|/¡ao/e ·e·/·|once Oene·o/ ·a/e· ´e/·m/c oc|/on· |o· /a//J/nq· bSl, ?004 13a Net|one| Annex to lurocode S bS lN !99S÷! bSl, ?00S 14 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS SS00÷! Conc·e|e - Como/emen|o·, S·/|/·/ ´|onJo·J |o S´ // ?06-' /e|/oJ o| ·oec/|,/nq onJ qa/Jonce |o |/e ·oec/|/e· bSl, ?00? 15 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS 4449 ´|ee/ |o· |/e ·e/n|o·cemen| o| conc·e|e - we/Jo//e ·e/n|o·c/nq ·|ee/ - So· co// onJ Jeco//eJ o·oJac| - ´oec/|/co|/on bSl, ?00S 16 CONS¹k0C¹ /o|/ono/ ·|·ac|a·o/ conc·e|e ·oec/|/co|/on |o· /a//J/nq con·|·ac|/on ¹h|rd ld|t|on, CS !S? ¹he Concrete Soc|ety on |ehe|¦ o¦ Construct, ?004 17 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !36/0 /\eca|/on o| conc·e|e ·|·ac|a·e· - /o·| ' Common bSl, |n preperet|on, due ?0!0 18 CONS¹k0C¹ /o|/ono/ ·|·ac|a·o/ conc·e|e ·oec/|/co|/on |o· /a//J/nq con·|·ac|/on lourth ed|t|on, |n preperet|on This publication is available for purchase from www.concretecentre.com 184 19 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS S666 ?00S, ´c/eJa//nq J/men·/on/nq /enJ/nq onJ ca||/nq o| ·|ee/ |o· ·e/n|o·cemen| ´oec/|/co|/on bSl, ?00S 20 ¹ll Q0llN'S lklN¹lk Ol AC¹S Ol lAkllAMlN¹, I/e Sa//J/nq /eqa/o|/on· ?000 ¹he Stet|onery O¦¦|ce l|m|ted, ?000 21 ¹ll S¹A¹lONlk¥ OlllCl llMl¹lL (¹SO), Sa//J/nq ¸/menJmen|j ¸/o ?j /eqa/o|/on· ?00? onJ |/e Sa//J/nq ¸/oo·o.eJ /n·oec|o·· e|cj ¸/menJmen|j /eqa/o|/on· ?00? ¹SO, ?00? 22 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON LL lNV !36/0÷! ?000 /\eca|/on o| conc·e|e ·|·ac|a·e· Common bSl, ?000 23 COOl, N /e·/qne·· qa/Je |o // '99'-4 /c|/on· on ·|·ac|a·e· w/nJ oc|/on· ¹homes ¹e|¦ord, london, ?00/ 24 LClC Oa/Je |o |/e a·e o| // '99'-'-4 - w/nJ oc|/on· LClC lu|||cet|ons, ?006 wwwcommun|t|esgovu|/p|enn|ngend|u||d|ng/p|enn|ng|u||d|ng/ |u||d|ngregu|et|onsreseerch/|u||d|ngd|v|s|onreseerch/ 25 bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON bS lN !99!÷!÷4 lurocode ! /c|/on· on ·|·ac|a·e· Oene·o/ oc|/on· w/nJ oc|/on· bSl, ?00S 25a Net|one| Annex to lurocode ! ÷ lert !÷4 bSl, ?00S 26 bkOOllk, O Conc·e|e /a//J/nq· ·c/eme Je·/qn monao/ (lC? ed|t|on) CCll-0S! ¹he Concrete Centre, ?009 27 ¹ll CONCkl¹l SOCll¹¥ ¹echn|ce| keport 64, Oa/Je |o |/e Je·/qn onJ con·|·ac|/on o| /C |/o| ·/o/·, CCll-0?? ¹he Concrete Soc|ety, ?00/ 28 COOLClllL, C l 8 vlbS¹lk, k M /C ·o·eoJ·/ee|· .3, CCll-00SCL c·e· qa/Je |o /C ·o·eoJ·/ee|· \3, CCll-00S ¹he Concrete Centre, ?006 29 lkvlN, A v /e·/qn o| ·/eo· uo// /a//J/nq· keport k!0? ClklA, !9S4 30 ¹ll CONCkl¹l ClN¹kl wo·/eJ e\omo/e· |o /a·ocoJe ? \o/ame ? CCll-04? ln preperet|on 31 MOSlll¥, b, b0NCl¥, ¦ 8 l0lSl, k /e/n|o·ceJ conc·e|e Je·/qn |o /a·ocoJe ?, S|xth ld|t|on le|greve McM|||en, ?00/ 32 ¹ll CONCkl¹l SOCll¹¥ /e|/ec|/on· /n conc·e|e ·/o/· onJ /eom· ¹k SS ¹he Concrete Soc|ety, ?00S 33 bllb¥, A /oJ/|/eJ o·ooo·o/· |o· con|·o///nq Je|/ec|/on· /, meon· o| ·o|/o· o| ·oon |o e||ec|/.e Jeo|/ ¹echn|ce| keport 4S6, Cement end Concrete Assoc|et|on, vexhem Spr|ngs, !9/! 34 AlllN, A l /e/n|o·ceJ conc·e|e Je·/qn |o S´ S''0 ·/mo/, e\o/o/neJ Spon, london, !9SS 35 lNS¹l¹0¹lON Ol S¹k0C¹0kAl lNClNllkS /onao/ |o· |/e Je·/qn o| conc·e|e /a//J/nq ·|·ac|a·e· |o /a·ocoJe ? lStructl, ?006 Further reading N bllb¥ A v 8 NAkA¥ANAN k S Les|gners gu|de to the lurocodes ÷ lN !99?÷!÷! end lN !99?÷!÷? lurocode ? /e·/qn o| conc·e|e ·|·ac|a·e· Oene·o/ ·a/e· onJ ·a/e· |o· /a//J/nq· onJ ·|·ac|a·o/ |/·e Je·/qn ¹homes ¹e|¦ord, ?00S N bkl¹lSl S¹ANLAkLS lNS¹l¹0¹lON ll !990?00/ ´|·ac|a·o/ /a·ocoJe· /\|·oc|· |·om |/e ·|·ac|a·o/ /a·ocoJe· |o· ·|aJen|· o| ·|·ac|a·o/ Je·/qn second ed|t|on bSl, london, ?00/ N llNL¥ C k 8 SMl¹l L A, Les|gners gu|de to the lurocodes ÷ lN !99?÷? lurocode ? /e·/qn o| conc·e|e ·|·ac|a·e· /o·| ? Conc·e|e /·/Jqe·, ¹homes ¹e|¦ord, ?00/ N ¹ll CONCkl¹l SOCll¹¥ ¹echn|ce| keport 64, Oa/Je |o |/e Je·/qn onJ con·|·ac|/on o| /C |/o| ·/o/·, CCll-0?? ¹he Concrete Soc|ety, ?00/ This publication is available for purchase from www.concretecentre.com 185 Append|x A Ler|ved ¦ormu|ee Appendix A: Derived formulae Flexure: beams and slabs Singly reinforced sections ¹he rectengu|er stress ||oc| shown |e|ow |n l|gure A! mey |e used h A s2 A s d a) Section b) Strain c) Forces neutral axis d 2 e c e sc lx e s x nf cd F c z F st F sc Figure A1 Strains and forces in a section Fig. 3.5 lor gredes o¦ concrete up to CS0/60, e cu 0003S, n ! end l 0S | cd a cc | c| /g C 0SS | c| /!S 3.1.6(1), 2.4.2.4(1) & NA | yd | y| / g S | y| /!!S 0S/| y| 2.4.2.4(1) & NA lor s|ng|y re|n¦orced sect|ons, the des|gn equet|ons cen |e der|ved es ¦o||ows Lever arm, z z = d – 0.4x F st F c 0.8x Figure A2 Beam lever arm / c (0SS | c| /!S) / (0S\) 04S3| c| /\ / st 0S// s | y| Cons|der moment ‡ , /, e|out the centre o¦ the tens|on ¦orce / 04S3| c| /\· Now · J ÷ 04\ = \ ?S(J ÷ ·) / 04S3| c| /?S(J ÷ ·)· !!333 (| c| /·J ÷ | c| / · ? ) let / ///J ? | c| !!333 (| c| /·J ÷ | c| /· ? )//J ? | c| !!333(·J ÷ · ? )/J ? = 0 !!333|(·/J) ? ÷ (·/J)| ÷ / (·/J) ? ÷ (·/J) ÷ 0SS?3S/ ‡ ln prect|ce the des|gn moment, / ld wou|d |e used A1 A1.1 This publication is available for purchase from www.concretecentre.com 186 So|v|ng the quedret|c equet|on ·/J |! ÷ (! ÷ 3S?9/) 0S |/? · J|! ÷ |! ÷ 3S?9/) 0S |/? Table C5 /| /· con·/Je·eJ qooJ o·oc|/ce /n |/e c/ |o //m/| ·/J |o o mo\/mam o| 09'J ¸I//· qao·J· oqo/n·| ·e/,/nq on .e·, |//n ·ec|/on· o| conc·e|e u//c/ o| |/e e\|·eme |oo o| o ·ec|/on mo, /e o| ¡ae·|/ono//e ·|·enq|/j Io//e· q/./nq .o/ae· o| ·/J onJ \/J |o· .o/ae· o| / mo, /e a·eJ Area of reinforcement, A s ¹e||ng moments e|out the centre o¦ the compress|on ¦orce / 0S// s | y| · / s //(0S/| y| ·) Limiting value of relative flexural compressive stress, K' Assum|ng no red|str||ut|on te|es p|ece, e ||m|t|ng ve|ue (on the strength o¦ concrete |n compress|on) ¦or / cen |e ce|cu|eted (denoted /) es ¦o||ows e cu3 concrete stre|n 0003S e s re|n¦orcement stre|n S00/(!!S × ?00 × !0 3 ) 000?? lrom stre|n d|egrem, l|gure A! \ 0003SJ/(0003S ÷ 000??) 06J lrom equet|ons e|ove / 04S3| c| /\· / 04S3| c| / 06J (J ÷ 04 × 06 J) 0?0/| c| / J ? = / 0?0/ lt |s o¦ten cons|dered good prect|ce to ||m|t the depth o¦ the neutre| ex|s to evo|d 'over- re|n¦orcement' (|e to ensure thet the re|n¦orcement |s y|e|d|ng et ¦e||ure, thus evo|d|ng |r|tt|e ¦e||ure o¦ the concrete) O¦ten \/J |s ||m|ted to 04S ¹h|s |s re¦erred to es the |e|enced sect|on |eceuse et the u|t|mete ||m|t stete the concrete end stee| reech the|r u|t|mete stre|ns et the seme t|me |3!| ¹h|s |s not e lurocode ? requ|rement end |s not eccepted |y e|| eng|neers Nonethe|ess ¦or \ 04SJ lrom equet|ons e|ove / 04S3| c| /\· / 04S3| c| / 04SJ (J ÷ 04 × 04SJ) 0!6/| c| /J ? = / 0!6/ Cl. 5.5(4) \/J |s e|so restr|cted |y the emount o¦ red|str||ut|on cerr|ed out lor | c| < S0 Mle d ~ 04 ÷ (06 ÷ 000!4e cu )\ u /J where J red|str||uted moment/e|est|c |end|ng moment |e¦ore red|str||ut|on \ u depth o¦ the neutre| ex|s et 0lS e¦ter red|str||ut|on e cu compress|ve stre|n |n the concrete et 0lS ¹h|s g|ves the ve|ues |n ¹e||e A! Table 3.1 This publication is available for purchase from www.concretecentre.com 187 Append|x A Ler|ved ¦ormu|ee A1.2 A2 A2.1 Table A1 Limits on D with respect to redistribution ratio, d d 1 0.95 0.9 0.85 0.8 0.75 0.7 % redistribution 0 S !0 !S ?0 ?S 30 K' 0?0S 0!9S 0!S? 0!6S 0!S3 0!3/ 0!?0 l¦ / > / the sect|on shou|d |e res|.ed or compress|on re|n¦orcement |s requ|red ln ||ne w|th cons|deret|on o¦ good prect|ce out||ned e|ove, this publication adopts a maximum value of K' = 0.167 Compression reinforcement, A s2 ¹he me|or|ty o¦ |eems used |n prect|ce ere s|ng|y re|n¦orced, end these |eems cen |e des|gned us|ng the ¦ormu|e der|ved e|ove ln some ceses, compress|on re|n¦orcement |s edded |n order to lncreese sect|on strength where sect|on d|mens|ons ere restr|cted, |e where N / > / ¹o reduce |ong term de¦|ect|on N ¹o decreese curveture/de¦ormet|on et u|t|mete ||m|t stete N A s2 A s Figure A3 Beam with compression reinforcement v|th re¦erence to l|gure A!, there |s now en extre ¦orce / sc 0S// s? | y| ¹he eree o¦ tens|on re|n¦orcement cen now |e cons|dered |n two perts, the ¦|rst pert to |e|ence the compress|ve ¦orce |n the concrete, the second pert to |e|ence the ¦orce |n the compress|on stee| ¹he eree o¦ tens|on re|n¦orcement requ|red |s there¦ore / s /' | cu /J ? /(0S/| y| ·) ÷ / s? where · |s ce|cu|eted us|ng /' |nsteed o¦ / / s? cen |e ce|cu|eted |y te||ng moments e|out the centre o¦ the tens|on ¦orce / / ÷ 0S/| y| / s? (J ÷ J ? ) / / | cu /J ? ÷ 0S/| y| / s? (J ÷ J ? ) keerreng|ng / s? (/ ÷ /) | c| /J ? /|0S/ | y| (J ÷ J ? )| Shear Shear resistance (without shear reinforcement), V Rd,c \ kd,c |C kd,c /(!00r! | c| ) !/3 ÷ / ! s cp | / w J ~ (. m|n ÷ / ! s cp ) / w J Exp. (6.2) where C kd,c 0!S/g C 0!S/!S 0!? NA / !÷ (?00/J) 0S < ?0 This publication is available for purchase from www.concretecentre.com 188 A2.2 A2.3 r ! / s /(/ w J) < 00? / ! 0!S s cp 0 ¦or non-prestressed concrete . m|n 003S/ !S | c| 0S =\ kd,c 0!?/(!00 r ! | c| ) !/3 / w J ~ 003S/ !S | c| 0S / w J Shear capacity Exp. (6.9) ¹he cepec|ty o¦ e concrete sect|on w|th vert|ce| sheer re|n¦orcement to ect es e strut, \ kd,mex \ kd,mex a cw / w ?v ! | cd /(cot y ÷ ten y) Cl. 6.2.3(3) Note 1, Exp. (6.6N) & NA where a cw !0 v ! v 06 |! ÷ | c| /?S0| | cd a cc | c| /g C !00 × | c| /!S =\ kd,mex 040 / w · | c| |! ÷ | c| /?S0|/(cot y ÷ ten y) keerreng|ng th|s equet|on g|ves y 0S s|n ÷! |. ld. /(0?0| c| |! ÷ | c| /?S0|)| ~ cot ÷! ?S where . ld. \ ld //· \ ld /(/09J) ln most ceses, where cot y ?S, y ?!S° . kd,mex,cot y ?S 0!3S/ w · | c| |! ÷ | c| /?S0| or . kd,mex,cot y ?S 0!3S| c| |! ÷ | c| /?S0| where . kd,mex,cot y ?S \ kd,mex,cot y ?S /(/·) \ kd,mex,cot y ?S /(09/J) vhere cot y > ?S, the eng|e o¦ the strut end . kd,mex shou|d |e ce|cu|eted, or . kd,mex mey |e |oo|ed up |n te||es or cherts (eg ¹e||e C/ or l|gure C!) Shear reinforcement Exp. (6.13) \ kd,s (/ sw /·)·| ywd (cot y ÷ cot a)s|n a ~ \ ld where / sw cross-sect|one| eree o¦ the sheer re|n¦orcement · spec|ng · |ever erm (epprox|mete ve|ue o¦ 09J mey norme||y |e used) | ywd | yw| /g S des|gn y|e|d strength o¦ the sheer re|n¦orcement a eng|e o¦ the ||n|s to the |ong|tud|ne| ex|s lor vert|ce| ||n|s, cot a 0 end s|n a !0 keerreng|ng ¦or vert|ce| ||n|s / sw /· ~ \ ld /·| ywd cot y or / sw /· ~ . ld,. / w / | ywd cot y M|n|mum eree o¦ sheer re|n¦orcement Exp. (9.5N) & NA / sw,m|n /(·/ w s|n a) ~ 00S| c| 0S /| y| where · |ong|tud|ne| spec|ng o¦ the sheer re|n¦orcement / w |reedth o¦ the we| This publication is available for purchase from www.concretecentre.com 189 Append|x A Ler|ved ¦ormu|ee A3 a eng|e o¦ the sheer re|n¦orcement to the |ong|tud|ne| ex|s o¦ the mem|er lor vert|ce| ||n|s s|n a !0 keerreng|ng ¦or vert|ce| ||n|s A sw,m|n /· ~ 00S/ w s|n a | c| 0S /| y| Columns a) Strain diagram b) Stress diagram n ex|s | cd a cc n| c| /g C J ? J c \ / J ? s sc s st e cu? e sc e y e c / s! / s? Figure A4 Section in axial compression and bending Fig. 6.1 lor ex|e| |oed / ld | cd /J c ÷ / s? s sc ÷ / s! s st but es / s? / s! / sN /? / ld | cd /J c ÷ / sN (s sc ÷ s st )/? / ld ÷ | cd /J c / sN (s sc ÷ s st )/? (/ ld ÷ | cd /J c )/(s sc ÷ s st ) / sN /? / sN /? (/ ld ÷ | cd /J c )/(s sc ÷ s st ) =/ sN /? (/ ld ÷ a cc n| c| /J c /g C )/(s sc ÷ s st ) lor moment e|out centre o¦ co|umn / ld | cd /J c (//? ÷ J c /?) ÷ / s? s sc (//? ÷ J ? ) ÷ / s! s st (//? ÷ J ? ) but es / s? / s! / sM /? / ld | cd /J c (//? ÷ J c /?) ÷ / sM (s sc ÷ s st )(//? ÷ J ? )/? / ld ÷ | cd /J c (//? ÷ J c /?) / sM (s sc ÷ s st )(//? ÷ J ? )/? |/ ld ÷ | cd /J c (//? ÷ J c /?)|/(s sc ÷ s st )(//? ÷ J ? ) / sM /? = / sM /? |/ ld ÷ a cc n| c| /J c (//? ÷ J c /?)/g C |/|(s sc ÷ s st )(//? ÷ J ? )| So|ut|on lterete \ such thet / sN / sM Note lor sect|ons who||y |n compress|on, the stre|n |s ||m|ted such thet everege stre|n ≤ e cs 000!/S (essum|ng ||||neer stress÷stre|n re|et|onsh|p) Cl. 6.1(6), Fig. 6.1, Table 3.1 This publication is available for purchase from www.concretecentre.com 190 B1 B1.1 B1.2 B1.3 Appendix B: Serviceability limit state Deflection ln meny ceses, pert|cu|er|y w|th s|e|s, de¦|ect|on |s cr|t|ce| to des|gn lurocode ?, C| /4 e||ows ¦or de¦|ect|on to |e contro||ed |y us|ng spendepth ret|o (//J) chec|s |n eccordence w|th C| /4? or |y ce|cu|et|on |n eccordence w|th C| /43 lt |s |mportent to d|¦¦erent|ete |etween the ver|ous methods used |n chec||ng de¦ormet|on es they w||| eech g|ve d|¦¦erent enswers ¹hree popu|er methods ere d|scussed |e|ow On|y thet descr||ed |n Sect|on b!! |e|ow |s su|te||e ¦or hend ce|cu|et|on TCC method [5,19] ¹he |n-serv|ce stress o¦ re|n¦orcement, s s , |s used to determ|ne e ¦ector, 3!0/s s , wh|ch |s used to mod|¦y the |es|c spen e¦¦ect|ve depth ret|o es e||owed |n C| /4?(?) o¦ lurocode ? |?| end modereted |y the Net|one| Annex |?e| ¹h|s method, h|gh||ghted es ¦ector l3 |n Conc/·e /a·ocoJe ? |S| , |s |ntended to |e used |n hend ce|cu|et|ons to der|ve (conservet|ve) ve|ues o¦ s s ¦rom eve||e||e 0lS moments ln eccordence w|th Note S o¦ ¹e||e NAS o¦ the 0l NA |?e| , the ret|o ¦or / s,prov // s,req |s restr|cted to !S |n e¦¦ect th|s ||m|ts the ¦ector 3!0/s s to !S where ‡ s s (| y| /g S ) (u qp /u u|t ) (/ s,req // s,prov ) /d < 3!0/!S where | y| cherecter|st|c strength o¦ re|n¦orcement S00 Mle g S pert|e| ¦ector ¦or re|n¦orcement !!S u qp ques|-permenent |oed (0Ll essumed) u perm u|t|mete |oed (0Ll essumed) / s,req eree o¦ re|n¦orcement requ|red / s,prov eree o¦ re|n¦orcement prov|ded d red|str||ut|on ret|o RC Spreadsheets method [28] ¹he kC spreedsheets ¹CCxxx|s |?S| use the spen depth method o¦ chec||ng de¦ormet|on |ut use en eccurete method ¦or determ|n|ng s s (see b3? |e|ow), wh|ch ege|n |s used to determ|ne the moderet|ng ¦ector 3!0/s s Age|n, |n eccordence w|th Note S o¦ ¹e||e NAS o¦ the 0l NA |?e| , the ret|o ¦or / s,prov // s,req |s restr|cted to !S |n e¦¦ect th|s ||m|ts the ¦ector 3!0/s s to !S Seperete ene|yses us|ng ques|-permenent |oeds need to |e cerr|ed out lor eech spen, en SlS neutre| ex|s depth |s determ|ned, then s c end s s ere der|ved ¦or the ques|-permenent |oed cond|t|ons ¹he ¦ector s s |s used |n eccordence w|th lurocode ? |?| end the current Net|one| Annex |?e| , to mod|¦y the |es|c spen e¦¦ect|ve depth ret|o vh||st th|s method g|ves e more eccurete end |ess conservet|ve essessment o¦ s s , |t |s on|y su|te||e ¦or computer spreedsheet epp||cet|ons See e|so Append|x bS ln the ene|ys|s o¦ s|e|s end |eems, supports ere usue||y essumed to |e p|nned ln ree||ty supports heve some cont|nu|ty, espec|e||y et end supports 0sue||y, nom|ne| top stee| |s essumed end prov|ded |n the top o¦ spens end |s used |n the determ|net|on o¦ sect|on propert|es Rigorous analysis k|gorous ene|ys|s, such es thet used |n the ser|es o¦ kC Spreedsheets ¹CCxxkx|s mey |e used to essess de¦ormet|on |n eccordence w|th lurocode ?, C| /43 ‡ See Append|x b!S This publication is available for purchase from www.concretecentre.com 191 Append|x b Serv|cee||||ty ||m|t stete ln the spreedsheets, sect|ons et !/?0th po|nts e|ong the |ength o¦ e spen ere chec|ed to determ|ne whether the ¦|exure| tens||e stress |n the sect|on |s |||e|y to exceed the tens||e strength o¦ the concrete dur|ng e|ther construct|on or serv|ce ||¦e seperete ene|yses ere underte|en us|ng ¦requent |oeds, ques|-permenent end temporery |oeds l¦ the ¦|exure| tens||e strength |s exceeded under ¦requent |oeds, then the sect|on |s essumed to |e crec|ed end reme|n crec|ed crec|ed sect|on propert|es ere used to determ|ne the red|us o¦ curveture ¦or thet !/?0th o¦ spen l¦ ¦|exure| tens||e strength |s not exceeded, un-crec|ed sect|on propert|es ere used ked|| o¦ curveture ere ce|cu|eted ¦or eech !/?0th spen |ncrement o¦ the e|ement us|ng the re|event propert|es end moments der|ved ¦rom ene|ys|s o¦ ques|-permenent ect|ons Le¦ormet|on |s ce|cu|eted ¦rom the |ncrements' curvetures v|e numer|ce| |ntegret|on over the |ength o¦ eech spen ¹he method |s |n eccordence w|th ¹he Concrete Soc|ety's pu|||cet|on ¹kSS |3?| Age|n the method |s su|te||e on|y ¦or computer epp||cet|ons end not ¦or hend ce|cu|et|on Differing results Lur|ng ?00S, |t |eceme |ncrees|ng|y epperent thet there ere |ncons|stenc|es |etween the resu|ts g|ven |y the r|gorous ce|cu|et|on method end spendepth methods descr||ed |n lurocode ? 0s|ng the r|gorous method g|ves de¦|ect|ons thet ere greeter then wou|d |e expected ¦rom the essumpt|ons steted ¦or //J methods |e de¦|ect|on ||m|ts o¦ //?S0 overe|| (see C| /4!(4)) or //S00 e¦ter construct|on (see C| /4!(S)) lt |s suspected thet th|s d|sper|ty |s the seme es thet exper|enced |etween spendepth end ce|cu|et|on methods |n bS S!!0 e d|sper|ty thet wes recogn|sed es |ong ego es !9/! |33| ¹he r|gorous method descr||ed e|ove re||es on meny essumpt|ons end |s |erge|y unce|||reted ege|nst ree| structures As noted |n ¹kSS, there |s en urgent need ¦or dete ¦rom ectue| structures so thet methods mey |e ce|||reted lt shou|d |e noted thet the r|gorous ene|ys|s method o|servet|ons were mede us|ng ¦requent |oeds where, |n eccordence w|th lurocode ?, ques|-permenent |oeds ere ce||ed ¦or lnd spens ere usue||y cr|t|ce| v|th respect to the r|gorous ene|ys|s method, |t hes |een suggested thet ¦or end-spens, the ¹CC end kC-spreedsheet methods resu|t |n de¦|ect|ons c|ose to the ||m|ts steted |n lurocode ?, prov|ded thet e nom|ne| end-support restre|n|ng moment |s present where none |s essumed |n ene|ys|s Ceut|on |s there¦ore necessery |n true p|nned end-support s|tuet|ons |ut where some cont|nu|ty ex|sts, th|s d|sper|ty mey |e eddressed |y ensur|ng thet eppropr|ete emounts o¦ re|n¦orcement, |n eccordence w|th the Code end Net|one| Annex, ere prov|ded et end supports ¹he NLl ¦or C| 9?!?(!) |n the 0l NA |?e| to bS lN !99?÷!÷? st|pu|etes thet ?S% o¦ end spen moment shou|d |e used to determ|ne end support re|n¦orcement ¹h|s |s usue||y eccommodeted |y prov|d|ng ?S% o¦ end spen |ottom stee| es top stee| et end supports lt |s on th|s |es|s thet the ce|cu|et|ons |n th|s pu|||cet|on ere cons|dered es |e|ng ¦urther su|stent|eted Note regarding factor 310/s s (factor F3) At the t|me o¦ pu|||cet|on (Lecem|er ?009) the euthors were ewere o¦ e pro|e||e chenge to 0l NA |?e| ¹e||e NAS wh|ch, |n e¦¦ect, wou|d meen thet the ¦ector 3!0/s s (l3) / s,prov // s,req ≤ !S, thus d|se||ow|ng the eccurete method out||ned |n Sect|ons 3!, 3?, 33, 34, 43 end Append|ces b!!, b!? end C/ Neutral axis at SLS ¹o ¦|nd \, neutre| ex|s, end serv|ces stresses s c end s s ¦or e concrete sect|on, et SlS, cons|der the crec|ed sect|on |n l|gure b! B1.4 B1.5 B2 This publication is available for purchase from www.concretecentre.com 192 d 2 d x A s2 A s a e – 1 a e E/E c Relative modulus Where a e = modular ratio E s /E c 1 Figure B1 Cracked concrete section at SLS lrom ¦|rst pr|nc|p|es, ¦or e ¦u||y crec|ed trens¦ormed sect|on, ¹ote| eree o¦ sect|on / /\ ÷ / s a e ÷ / s? (a e ÷ !) !st moment o¦ eree, / y /\ ? /? ÷ / s Ja e ÷ / s? J ? (a e ÷ !) lor e s|e|, / !000, there¦ore / !000\ ÷ / s a e ÷ / s? (a e ÷ !) / y S00\ ? ÷ / s Ja e ÷ / s? J ? (a e ÷ !) Neutre| ex|s depth, \ \ / y // |S00\ ? ÷ / s Ja e ÷ / s? J ? (a e ÷ !)|/|!000\ ÷ / s a e ÷ / s? (a e ÷ !)| ¹here¦ore \|!000\ ÷ / s a e ÷ / s? (a e ÷ !)| |S00\ ? ÷ / s Ja e ÷ / s? J ? (a e ÷ !)| 0 |S00\ ? ÷ / s Ja e ÷ / s? J ? (a e ÷ !)| ÷ \|!000\ ÷ / s a e ÷ / s? (a e ÷ !)| S00\ ? ÷ \|!000\| ÷ / s Ja e ÷ / s? J ? (a e ÷ !)| ÷ \|/ s a e ÷ / s? (a e ÷ !)| ÷ S00\ ? ÷ \|/ s a e ÷ / s? (a e ÷ !)| ÷ |/ s Ja e ÷ / s? J ? (a e ÷ !)| So|v|ng the quedret|c \ ÷/ ± / ? ÷ 4ac) 0S |/?o \ ÷|/ s a e ÷ / s? (a e ÷ !) ± [|/ s a e ÷ / s? (a e ÷ !)| ? ÷ 4 × S00 × |/ s Ja e ÷ / s? J ? (a e ÷ !)|¦ 0S | (? x S00) or trenspos|ng, \ |÷(a e ÷ !)/ s? ÷ a e / s ÷ [|(a e ÷ !)/ s? ÷ a e / s | ? ÷ ?000|(a e ÷!)/ s? J ? ÷ a e / s J|¦ 0S | !000 or \ |÷(a e ÷ !)/ s? ÷ a e / s ÷ [|(a e ÷ !)/ s? ÷ a e / s | ? ÷ ?/|(a e ÷ !)/ s? J ? ÷ a e / s J|¦ 0S | / ¹h|s express|on |s used |n the kC spreedsheets |?S| This publication is available for purchase from www.concretecentre.com 193 Append|x b Serv|cee||||ty ||m|t stete SLS stresses in concrete, s c , and reinforcement, s s Singly reinforced section Cons|der the s|ng|y re|n¦orced sect|on |n l|gure b? a) Section b) Dimensions and forces M qp A s F c F s x/ 3 x z b Figure B2 SLS stresses: singly reinforced section Cons|der moments e|out / c / qp / s · / s (J ÷ \/3) / s / qp /(J ÷ \/3) s s / qp /|/ s (J ÷ \/3)| s s / s / qp /(J ÷ \/3) \/s c /? s c ?s s / s /\/ Doubly reinforced section Cons|der the s|ng|y re|n¦orced sect|on |n l|gure b3 a) Section b) Dimensions, modular ratios and stresses M qp d 2 A s A s2 E/E c = a e – 1 s = s c (a e – 1)(x – d 2 )/x E/E c = a e s = s c E/E c = a e s = s s x d b Figure B3 SLS stresses Cons|der moment, / qp , e|out |ottom re|n¦orcement, / s |34| / qp / s? (J ÷ J ? )(a e ÷ !)[(\ ÷ J ? )/\¦ s c ÷ s c /(\/?)(J ÷ \/3) ¹here¦ore s c / qp / |/ s? (J ÷ J ? )(a e ÷ !) )[(\ ÷ J ? )/\¦ ÷ /(\/?)(J ÷ \/3)| And ¦rom stress d|egrem s s s c a e (J ÷ \)/\ B3 B3.1 B3.2 This publication is available for purchase from www.concretecentre.com 194 Appendix C: Design aids ¹he ¦o||ow|ng te||es, text end ¦|gures heve |een der|ved ¦rom lurocode ? end ere prov|ded es des|gn e|ds ¦or des|gners |n the 0l ¹hese des|gn e|ds heve |een re¦erenced |n the text end genere||y heve |een te|en ¦rom Sect|on !S o¦ Conc/·e /a·ocoJe ? |S| Design values of actions lor the 0lS o¦ strength (S¹k) where there |s e s|ng|e ver|e||e ect|on use e|ther !3S N O | ÷ !SÇ | lxp (6!0) ¦rom bS lN !990 |!0| or the worse cese o¦ !3S N O | ÷ c 0 !SÇ | lxp (6!0e) end !?S N O | ÷ !SÇ | lxp (6!0|) where c 0 !0 ¦or storege, 0S ¦or snow |ut otherw|se 0/, see ¹e||e ?? ln most ceses lxp (6!0|) w||| |e eppropr|ete, except ¦or storege where the use o¦ lxp (6!0e) |s |||e|y to |e more onerous lor the SlS o¦ de¦ormet|on, ques|-permenent |oeds shou|d |e epp||ed ¹hese ere !0O | ÷ c ? Ç | where c ? |s dependent on use, eg 03 ¦or o¦¦|ces end res|dent|e| end 0/ ¦or storege Values of actions ¹he ve|ues o¦ ect|ons (|e |oeds) ere de¦|ned |n lurocode ! |!!| ¹he perts o¦ lurocode ! ere g|ven |n ¹e||e C! ¹hese ve|ues ere te|en es cherecter|st|c ve|ues At the t|me o¦ pu|||cet|on, the 0l Net|one| Annexes to these perts ere |n ver|ous stetes o¦ reed|ness As lL 66S/ |6| me|es c|eer, unt|| the eppropr|ete luropeen stenderds |ecome eve||e||e, des|gners mey cons|der us|ng current prect|ce or current br|t|sh Stenderds |n con|unct|on w|th lurocode ?, prov|ded they ere compet|||e w|th lurocode ? end thet the resu|t|ng re||e||||ty |s eccepte||e bS lN !99!÷!÷! stetes thet the dens|ty o¦ concrete |s ?4 |N/m 3 , re|n¦orced concrete, ?S |N/m 3 end wet re|n¦orced concrete, ?6 |N/m 3 Table C1 The parts of Eurocode 1 [11] Reference Title BS EN 1991-1-1 Lens|t|es, se|¦-we|ght end |mposed |oeds BS EN 1991-1-2 Act|ons on structures exposed to ¦|re BS EN 1991-1-3 Snow |oeds BS EN 1991-1-4 v|nd ect|ons BS EN 1991-1-5 ¹herme| ect|ons BS EN 1991-1-6 Act|ons dur|ng execut|on BS EN 1991-1-7 Acc|dente| ect|ons due to |mpect end exp|os|ons BS EN 1991-2 ¹re¦¦|c |oeds on |r|dges BS EN 1991-3 Act|ons |nduced |y crenes end mech|nery BS EN 1991-4 Act|ons |n s||os end ten|s C1 C2 This publication is available for purchase from www.concretecentre.com 195 Append|x C Les|gn e|ds Analysis Ane|ys|s |s dee|t w|th |n Sect|on S o¦ Conc/·e /a·ocoJe ? vhere eppropr|ete the coe¦¦|c|ents g|ven |n ¹e||es C? end C3 cen |e used to determ|ne des|gn moments end sheer ¦or s|e|s end |eems et 0lS Table C2 Coefficients for use with one-way spanning slabs to Eurocode 2 Coefficient Location End support/slab connection Internal supports and spans Pinned end support Continuous Outer support Near middle of end span Outer support Near middle of end span At 1st interior support At middle of interior spans At interior supports Moment 00 00S6 ÷ 004 00/S ÷ 00S6 0063 ÷ 0063 Shear 040 ÷ 046 ÷ 060060 ÷ 0S00S0 Notes 1 App||ce||e to one-wey spenn|ng s|e|s where the eree o¦ eech |ey exceeds 30 m ? , Ç | < !?SO | end ¡ | < S |N/m ? , su|stent|e||y un|¦orm |oed|ng (et |eest 3 spens, m|n|mum spen ~ 0SS mex|mum (des|gn) spen 2 Les|gn moment coe¦¦ x n x spen ? end des|gn sheer coe¦¦ x n x spen where n |s e 0Ll w|th e s|ng|e ver|e||e ect|on g C q | ÷ cg Q ¡ | where q | end ¡ | ere cherecter|st|c permenent end ver|e||e ect|ons |n |N/m 3 bes|s ¥|e|d ||ne des|gn (essumed ?0% red|str||ut|on |/| ) Table C3 Coefficients for use with beams (and one-way spanning slabs) to Eurocode 2 Coefficient Location Outer support Near middle of end span At 1st interior support At middle of interior spans At interior supports Moment ` k and j k ?S% spen a ÷ 0094 ÷ 00/S Moment ` k ÷ 0090 ÷ 0066 ÷ Moment j k ÷ 0!00 ÷ 00S6 ÷ Shear 04S ÷ 0630SS ÷ 0S00S0 b Notes 1 lor |eems end s|e|s, 3 or more spens (¹hey mey e|so |e used ¦or ?-spen |eems |ut support moment coe¦¦|c|ent 0!06 end |nterne| sheer coe¦¦|c|ent 063 |oth s|des) 2 Cenere||y Ç | < O | , end the |oed|ng shou|d |e su|stent|e||y un|¦orm|y d|str||uted Otherw|se spec|e| curte||ment o¦ re|n¦orcement |s requ|red 3 M|n|mum spen ~ 0SS x mex|mum (end des|gn) spen 4 Les|gn moment et supports coe¦¦ x n x spen ? or |n spens (coe¦¦ q | x g C q | ÷ coe¦¦ ¡ | x cg Q ¡ | ) x spen ? 5 Les|gn sheer et centre||ne o¦ supports coe¦¦ x n x spen where n |s e 0Ll w|th e s|ng|e ver|e||e ect|on g C q | ÷ cg Q ¡ | where q | end ¡ | ere cherecter|st|c permenent end ver|e||e ect|ons |n |N/m g C end cg Q ere dependent on use o¦ bS lN !990, lxpress|ons (6!0), (6!0e) or (6!0|) See Sect|on C! 6 bes|s A||- end e|ternete-spens-|oeded ceses es 0l Net|one| Annex end !S% red|str||ut|on et supports Key a At outer support '?S% spen' re|etes to the 0l Net|one||y Leterm|ned leremeter ¦or lurocode ?, C| 9?!?(!) ¦or m|n|mum percentege o¦ spen |end|ng moment to |e essumed et supports |n |eems |n mono||th|c construct|on !S% mey |e eppropr|ete ¦or s|e|s (see lurocode ?, C| 93!?) b lor |eems o¦ ¦|ve spens, 0SS epp||es to centre spen Cl. 9.2.1.2 Cl. 9.3.1.2 C3 This publication is available for purchase from www.concretecentre.com 196 Design for bending Leterm|ne whether N / < / or not (|e whether under-re|n¦orced or not) where / / ld /(/J ? | c| ) where J e¦¦ect|ve depth / ÷ cover ÷ f/? / w|dth o¦ sect|on |n compress|on / mey |e determ|ned ¦rom ¹e||e C4 end |s dependent on the red|str||ut|on ret|o used Table C4 Values for D Redistribution ratio, d s/] for D a D a 1 – d 1.00 0/6 (0S?) 0?0S (0!6S) 0% 0.95 0/S (0S?) 0!9S (0!6S) S% 0.90 0S0 (0S?) 0!S? (0!6S) !0% 0.85 0S? 0!6S !S% 0.80 0S4 0!S3 ?0% 0.75 0S6 0!3/ ?S% 0.70 0SS 0!?0 30% Note C|ess A re|n¦orcement |s restr|cted to e red|str||ut|on ret|o, d < 0S Key a lt |s recommended thet \/J |s ||m|ted to 04S |3S| As e consequence ·/J |s ||m|ted to e m|n|mum o¦ 0S?0 end / to e m|n|mum o¦ 0!6S l¦ N / < /, sect|on |s under-reinforced lor rectengu|er sect|ons / s! / ld /| yd · where / s! eree o¦ tens||e re|n¦orcement / ld des|gn moment | yd | y| /g S S00/!!S 434S Mle · J|0S ÷ 0S(! ÷ 3S3/) 0S | < 09SJ Ve|ues o¦ ·/J (end \/J) mey |e te|en ¦rom ¹e||e CS lor ¦|enged |eems where x < !?S/ ¦ , / s! / ld /| yd · where \ depth to neutre| ex|s Ve|ues o¦ \/J mey |e te|en ¦rom ¹e||e CS / ¦ th|c|ness o¦ ¦|enge How to: Beams [8] lor ¦|enged |eems where \ ~ !?S/ ¦ , re¦er to /ou |o Je·/qn conc·e|e ·|·ac|a·e· a·/nq /a·ocoJe ? |S| l¦ N / > /, sect|on |s over-reinforced end requ|res compress|on re|n¦orcement / s? (/ ld ÷ /)/| sc (J ÷ J ? ) where / s? compress|on re|n¦orcement l¦ J ? /x > 03/S then the term / s? shou|d |e rep|eced |y the term !6(! ÷ J ? /\) / s? / //J ? | c| | sc /00(\ u ÷ J ? )/\ u < | yd where J ? e¦¦ect|ve depth to compress|on re|n¦orcement C4 This publication is available for purchase from www.concretecentre.com 197 Append|x C Les|gn e|ds \ u (d ÷ 04)J where d red|str||ut|on ret|o ¹ote| eree o¦ stee| / s! //(| yd ·) ÷ / s? | sc /| yd Table C5 Values of s/] and q/] for singly reinforced rectangular sections D s/] q/] (1 – d) max a 0.04 09S0 b 0!?S 30% 0.05 09S0 b 0!?S 30% 0.06 0944 0!40 30% 0.07 0934 0!6S 30% 0.08 09?4 0!9! 30% 0.09 09!3 0?!/ 30% 0.10 090? 0?4S 30% 0.11 0S9! 0?/? 30% 0.12 0SS0 030! 30% 0.13 0S6S 033! ?/% 0.14 0SS6 036! ?4% 0.15 0S43 0393 ?!% 0.16 0S30 04?S !S% 0.17 0S!6 c 0460 c !4% 0.18 0S0? c 049S c !!% 0.19 0/S/ c 0S33 c /% 0.20 0//! c 0S/? c 3% 0.208 0/SS c 0606 c 0% Note | c| < S0 Mle Key a Mex|mum e||owe||e red|str||ut|on b lrect|ce| ||m|t c lt |s recommended thet \/J |s ||m|ted to 04S0 |3S| As e consequence ·/J |s ||m|ted to e m|n|mum o¦ 0S?0 end / to 0!6S Design for beam shear Requirement for shear reinforcement l¦ . ld > . kd,c then sheer re|n¦orcement |s requ|red where . ld \ ld // w J, ¦or sect|ons w|thout sheer re|n¦orcement (|e s|e|s) . kd,c sheer res|stence w|thout sheer re|n¦orcement, ¦rom ¹e||e C6 C5 C5.1 This publication is available for purchase from www.concretecentre.com 198 C5.2 Table C6 Shear resistance without shear reinforcement, o Rd,c (MPa) r l = : sl / [ w ] Effective depth ] (mm) G 200 225 250 275 300 350 400 450 500 600 750 < 0.25% 0S4 0S? 0S0 04S 04/ 04S 043 04! 040 03S 036 0.50% 0S9 0S/ 0S6 0SS 0S4 0S? 0S! 049 04S 04/ 04S 0.75% 06S 066 064 063 06? 0S9 0SS 0S6 0SS 0S3 0S! 1.00% 0/S 0/? 0/! 069 06S 06S 064 06? 06! 0S9 0S/ 1.25% 0S0 0/S 0/6 0/4 0/3 0/! 069 06/ 066 063 06! 1.50% 0SS 0S3 0S! 0/9 0/S 0/S 0/3 0/! 0/0 06/ 06S 1.75% 090 0S/ 0SS 0S3 0S? 0/9 0// 0/S 0/3 0/! 06S ~ 2.00% 094 09! 0S9 0S/ 0SS 0S? 0S0 0/S 0// 0/4 0/! Notes 1 ¹e||e der|ved ¦rom lurocode ? end 0l Net|one| Annex 2 ¹e||e creeted ¦or | c| 30 Mle essum|ng vert|ce| ||n|s 3 lor r | ~ 04% end | c| ?S Mle, epp|y ¦ector o¦ 094 | c| 40 Mle, epp|y ¦ector o¦ !!0 | c| S0 Mle, epp|y ¦ector o¦ !!9 | c| 3S Mle, epp|y ¦ector o¦ !0S | c| 4S Mle, epp|y ¦ector o¦ !!4 Not epp||ce||e ¦or | c| > S0 Mle Section capacity check l¦ . ld,. > . kd,mex then sect|on s|.e |s |nedequete where . ld,. \ ld // w · \ ld // w 09J, ¦or sect|ons u/|/ sheer re|n¦orcement . kd,mex cepec|ty o¦ concrete struts expressed es e stress |n the vert|ce| p|ene \ kd,mex // w · \ kd,mex // w 09J . kd,mex cen |e determ|ned ¦rom ¹e||e C/, |n|t|e||y chec||ng et cot y ?S Shou|d |t |e requ|red, e greeter res|stence mey |e essumed |y us|ng e |erger strut eng|e, y Table C7 Capacity of concrete struts expressed as a stress, o Rd,max _ ck o Rd,max (MPa) Strength reduction factor, v cot y 2.50 2.14 1.73 1.43 1.19 1.00 y 2.18° 25° 30° 35° 40° 45° 20 ?S4 ?S? 3!9 346 36? 36S 0SS? 25 3!0 34S 390 4?3 443 4S0 0S40 30 364 404 4S/ 496 S?0 S?S 0S?S 35 4!S 46! S?! S66 S93 60? 0S!6 40 463 S!S SS? 63! 66? 6/? 0S04 45 S09 S6S 639 693 /?/ /3S 049? 50 SS? 6!3 693 /S? /SS S00 04S0 Notes 1 ¹e||e der|ved ¦rom lurocode ? end 0l Net|one| Annex essum|ng vert|ce| ||n|s, |e cot a 0 2 v 06|! ÷ (| c| /?S0)| 3 . kd,mex v| cd (cot y ÷ cot a)/(! ÷ cot ? y) This publication is available for purchase from www.concretecentre.com 199 Append|x C Les|gn e|ds C5.3 Shear reinforcement design / sw /· ~ . ld,. / w /| ywd cot y where / sw eree o¦ sheer re|n¦orcement (vert|ce| ||n|s essumed) · spec|ng o¦ sheer re|n¦orcement . ld,. \ ld // w ·, es |e¦ore / w |reedth o¦ the we| | ywd | yw| /g S des|gn y|e|d strength o¦ sheer re|n¦orcement Cenere||y / sw /· ~ . ld,. / w /!0S/ where | yw| S00 Mle, g S !!S end cot y ?S A|ternet|ve|y, / sw /· per metre w|dth o¦ / w mey |e determ|ned ¦rom l|gure C!e) or C!|) es |nd|ceted |y the ||ue errows |n l|gure C!e) ¹hese ¦|gures mey e|so |e used to est|mete the ve|ue o¦ cot y beems ere su||ect to e m|n|mum sheer ||n| prov|s|on Assum|ng vert|ce| ||n|s, / sw,m|n /·/ w ~ 00S | c| 0S /| y| (see ¹e||e CS) Table C8 Values of : sw,min /l[ w for beams for vertical links and _ yk = 500 MPa and compatible resistance, o Rd Concrete class C20/25 C25/30 C30/37 C35/45 C40/50 C45/55 C50/60 A sw,min /sb w for beams (x 10 3 ) 0/? 0S0 0SS 09S !0! !0/ !!3 v Rd for A sw,min /sb w (MPa) 0/S 0S/ 09S !03 !!0 !!/ !?3 4.0 3.0 2.0 5.0 6.0 7.0 8.0 0 2 4 6 8 10 12 14 C35/45 C40/50 C45/55 0.0 1.0 2.14 1.73 1.43 1.19 1.00 See Fig. C1b) A sw /s required per metre width of b w f ywk = 500 MPa v Rd,max for cot y = 2.5 v E d , z ( M P a ) C30/37 C20/25 C25/30 C50/60 Figure C1a) Diagram to determine : sw /l required (for beams with high shear stress) This publication is available for purchase from www.concretecentre.com 200 C6 Design for punching shear Leterm|ne |¦ punch|ng sheer re|n¦orcement |s requ|red, |n|t|e||y et a ! , then |¦ necessery et su|sequent per|meters, a | l¦ . ld > . kd,c then punch|ng sheer re|n¦orcement |s requ|red where . ld b\ ld /a | J where b ¦ector dee||ng w|th eccentr|c|ty \ ld epp||ed sheer ¦orce a | |ength o¦ the per|meter under cons|deret|on J meen e¦¦ect|ve depth . kd,c sheer res|stence w|thout sheer re|n¦orcement (see ¹e||e C6) lor vert|ce| sheer re|n¦orcement (/ sw /· r ) a ! (. ld ÷ 0/S . kd,c )/(!S | ywd,e¦ ) where / sw eree o¦ sheer re|n¦orcement |n one per|meter eround the co|umn lor / sw,m|n see Conc/·e /a·ocoJe ?, Sect|on !04? end ¦or |eyout see Sect|on !?43 · r red|e| spec|ng o¦ per|meters o¦ sheer re|n¦orcement Concise: 10.4.2, 12.4.3 a ! |es|c contro| per|meter ?J ¦rom co|umn ¦ece | ywd,e¦ e¦¦ect|ve des|gn strength o¦ re|n¦orcement (?S0 ÷ 0?SJ) < | ywd lor Crede S00 sheer re|n¦orcement see ¹e||e C9 Table C9 Values of _ ywd,ef for grade 500 reinforcement ] 150 200 250 300 350 400 450 f ywd,ef ?S/S 300 3!?S 3?S 33/S 3S0 36?S At the co|umn per|meter, chec| . ld ≤ \ kdmex ¦or cot y !0 g|ven |n ¹e||e C/ C20/25 C25/30 C30/37 C30/37 C35/45 C40/50 C45/55 C50/60 4.0 3.0 2.0 1.0 0.0 0 1 2 3 4 f ywk = 500 MPa A sw,min /s for beams Range of v Rd,c for range d = 200 mm, r = 2.0% to d = 750 mm, r = 0.5% C 2 5 /3 0 A sw /s required per metre width of b w v E d , z ( M P a ) C20/25 Figure C1b) Diagram to determine : sw /l required (for slabs and beams with low shear stress) This publication is available for purchase from www.concretecentre.com 201 Append|x C Les|gn e|ds C7 Check deflection ln genere|, the SlS stete o¦ de¦|ect|on mey |e chec|ed |y us|ng the spen-to-e¦¦ect|ve-depth epproech More cr|t|ce| eppre|se| o¦ de¦ormet|on |s outs|de the scope o¦ th|s pu|||cet|on ¹o use the spen-to-e¦¦ect|ve-depth epproech, ver|¦y thet A||owe||e //J / x / x l! x l? x l3 ~ ectue| //J where / |es|c spen-to-e¦¦ect|ve-depth ret|o der|ved ¦or / !0 end r' 0 ¦rom Sect|on !0S? o¦ Conc/·e /a·ocoJe ? or ¹e||e C!0 or l|gure C? Concise: 10.5.2 / ¦ector to eccount ¦or structure| system See ¹e||e C!! l! ¦ector to eccount ¦or ¦|enged sect|ons vhen / e¦¦ // w !0, ¦ector l! !0 vhen / e¦¦ // w |s greeter then 30, ¦ector l! 0S0 lor ve|ues o¦ / e¦¦ // w |etween !0 end 30, |nterpo|et|on mey |e used (see ¹e||e C!?) where / e¦¦ |s de¦|ned |n Sect|on S?? o¦ Conc/·e /a·ocoJe ? / w w|dth o¦ we| ln l |eems / w m|n|mum w|dth o¦ we| |n tens||e eree ln tepered we|s / w w|dth o¦ we| et centro|d o¦ re|n¦orcement |n we| Concise: 5.2.2 l? ¦ector to eccount ¦or |r|tt|e pert|t|ons |n essoc|et|on w|th |ong spens Cenere||y l? !0 |ut |¦ |r|tt|e pert|t|ons ere ||e||e to |e demeged |y excess|ve de¦|ect|on, l? shou|d |e determ|ned es ¦o||ows e) |n ¦|et s|e|s |n wh|ch the |onger spen |s greeter then SS m, l? SS// e¦¦ |) |n |eems end other s|e|s w|th spens |n excess o¦ /0 m, l? /0// e¦¦ Ve|ues o¦ l? mey |e te|en ¦rom ¹e||e C!3 l3 ¦ector to eccount ¦or serv|ce stress |n tens||e re|n¦orcement 3!0/s s < !S Conservet|ve|y, |¦ e serv|ce stress, s s , o¦ 3!0 Mle |s essumed ¦or the des|gned eree o¦ re|n¦orcement, / s,req then l3 / s,prov // s,req < !S More eccurete|y, ‡ the serv|cee||||ty stress, s s , mey |e est|meted es ¦o||ows s s | y| /g S |(O | ÷ c ? Ç | )/(!?SO | ÷ !SÇ | )| |/ s,req // s,prov | (!/d) or s s s su |/ s,req // s,prov | (!/d) where s su the unmod|¦|ed SlS stee| stress, te||ng eccount o¦ g M ¦or re|n¦orcement end o¦ go|ng ¦rom u|t|mete ect|ons to serv|cee||||ty ect|ons S00/g S (O | ÷ c ? Ç | )/(!?SO | ÷ !SÇ | ) s su mey |e est|meted ¦rom l|gure C3 es |nd|ceted |y the ||ue errow / s,req // s,prov eree o¦ stee| requ|red d|v|ded |y eree o¦ stee| prov|ded (!/d) ¦ector to 'un-red|str||ute' 0lS moments so they mey |e used |n th|s SlS ver|¦|cet|on (see ¹e||e C!4) Actue| //J ectue| spen d|v|ded |y e¦¦ect|ve depth, J ‡ See Append|x b!S This publication is available for purchase from www.concretecentre.com 202 Table 7.4N Table C10 Basic ratios of span-to-effective-depth, G, for members without axial compression Required reinforcement, r _ ck 20 25 30 35 40 45 50 0.30% ?S9 3?? 39? 466 S46 630 /!S 0.40% !9! ??4 ?6? 304 3S0 39S 4S0 0.50% !/0 !SS ?0S ?30 ?SS ?SS 3?0 0.60% !60 !/3 !SS !9S ?!3 ?3! ?S? 0.70% !S3 !64 !/4 !SS !96 ?06 ?!/ 0.80% !4S !S/ !66 !/6 !SS !94 ?04 0.90% !43 !S? !60 !6S !// !SS !93 1.00% !40 !4S !SS !63 !/0 !/S !SS 1.20% !3S !4! !4S !S4 !60 !66 !/3 1.40% !3! !3/ !4? !4S !S3 !SS !64 1.60% !?9 !33 !3S !43 !4S !S? !S/ 1.80% !?/ !3! !3S !39 !43 !4S !S? 2.00% !?S !?9 !33 !36 !40 !44 !4S 2.50% !?? !?S !?S !3! !34 !3/ !40 3.00% !?0 !?3 !?S !?S !30 !33 !3S 3.50% !!9 !?! !?3 !?S !?/ !?9 !3! 4.00% !!S !!9 !?! !?3 !?S !?/ !?9 4.50% !!/ !!S !?0 !?? !?3 !?S !?/ 5.00% !!6 !!S !!9 !?! !?? !?4 !?S Reference reinforcement ratio, r 0 04S% 0S0% 0SS% 0S9% 063% 06/% 0/!% Notes 1 vhere r / s //J 2 lor ¹-sect|ons r |s the eree o¦ re|n¦orcement d|v|ded |y the eree o¦ concrete e|ove the centro|d o¦ the tens|on re|n¦orcement 3 ¹he ve|ues ¦or spen-to-e¦¦ect|ve-depth heve |een |esed on ¹e||e /4N |n lurocode ?, us|ng / ! (s|mp|y supported) end r 0 (no compress|on re|n¦orcement requ|red) 4 ¹he spen-to-e¦¦ect|ve-depth ret|o shou|d |e |esed on the shorter spen |n two-wey spenn|ng s|e|s end the |onger spen |n ¦|et s|e|s 30 28 26 24 22 20 18 16 14 32 12 0.40% 0.60% 0.80% 1.00% 1.20% 1.40% 1.60% 1.80% 2.00% Design tension reinforcement (100A s,req /bd) B a s i c s p a n - t o - e f f e c t i v e - d e p t h r a t i o N ( l / d ) f ck = 50 f ck = 45 f ck = 40 f ck = 35 f ck = 30 f ck = 25 f ck = 20 Figure C2 Basic span-to-effective depth ratios, G, for D = 1, r' = 0 This publication is available for purchase from www.concretecentre.com 203 Append|x C Les|gn e|ds Table C11 D factors to be applied to basic ratios of span-to-effective-depth Structural system D Beams Slabs S|mp|y supported |eems One- or two-wey spenn|ng s|mp|y supported s|e|s !0 lnd spen o¦ cont|nuous |eems lnd spen o¦ one-wey spenn|ng cont|nuous s|e|s, or two-wey spenn|ng s|e|s cont|nuous over one |ong edge !3 lnter|or spens o¦ cont|nuous |eems lnter|or spens o¦ cont|nuous s|e|s !S ÷ l|et s|e|s (|esed on |onger spen) !? Cent||evers Cent||ever 04 Table C12 Factor F1, modifier for flanged beams [ eff /[ w 1.0 1.5 2.0 2.5 ~ 3.0 Factor !00 09S 090 0SS 0S0 Table C13 Factor F2, modifier for long spans supporting brittle partitions Span, m e eff < 7.0 7.5 8.0 8.5 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 Flat slabs 8.5/e eff !00 !00 !00 !00 094 0SS 0// 0/! 06S 06! 0S/ 0S3 Beams and other slabs 7.0/e eff !00 093 0SS 0S? 0/S 0/0 064 0SS 0S4 0S0 04/ 044 320 300 280 260 240 220 200 180 1.00 2.00 3.00 4.00 U n m o d i f i e d s t e e l s t r e s s s s u Ratio G k /Q k y 2 = 0 .6 , g G = 1 .2 5 y2 = 0 .3 , g G = 1 .2 5 y 2 = 0 .2 , g G = 1 .2 5 y 2 = 0.8, g G = 1.35 y 2 = 0.6, g G = 1.35 y 2 = 0 .3 , g G = 1 .3 5 y 2 = 0.2, g G = 1.35 Figure C3 Determination of unmodified SLS, stress in reinforcement, s su This publication is available for purchase from www.concretecentre.com 204 C8 Table C14 (1/d) factor to be applied to unmodified s su to allow for redistribution used Average redistribution used 20% 15% 10% 5% 0% –5% –10% –15% –20% –25% –30% Redistribution ratio used, d 1.20 1.15 1.10 1.05 1.00 0.95 0.90 0.85 0.80 0.75 0.70 (1/d) S3% S/% 9!% 9S% !00% !0S% !!!% !!S% !?S% !33% !43% Notes 1 vhere coe¦¦|c|ents ¦rom ¹e||e C? heve |een used |n des|gn end where Ç | &!?SO | , the coe¦¦|c|ents |n ¹e||e C? mey |e cons|dered to represent moment d|str||ut|on o¦ ÷S% neer m|dd|e o¦ end spen w|th p|nned end support ÷??% et ¦|rst |nter|or support, es e worst cese ÷3% neer m|dd|e o¦ |nterne| spens, es e worst cese ÷?S% et |nter|or supports, es e worst cese 2 vhere coe¦¦|c|ents ¦rom ¹e||e C3 heve |een used |n des|gn end where Ç | &O | , the coe¦¦|c|ents |n ¹e||e C3 mey |e cons|dered to represent moment red|str||ut|on o¦ ÷3% neer m|dd|e o¦ end spen w|th p|nned end support, es e worst cese ÷9% neer m|dd|e o¦ |nterne| spens, es e worst cese ÷!S% et e|| |nter|or supports Control of cracking Crec||ng mey |e contro||ed |y restr|ct|ng e|ther mex|mum |er d|emeter or mex|mum |er spec|ng to the re|event d|emeters end spec|ngs g|ven |n ¹e||e C!S ¹he eppropr|ete SlS stress |n re|n¦orcement, s s , mey |e determ|ned es out||ned ¦or l3 |n Sect|on C/ M|n|mum erees end espects o¦ dete|||ng shou|d |e chec|ed Table C15 Maximum bar diameters f or maximum bar spacing for crack control Steel stress (MPa) s s Maximum bar size (mm) OR Maximum bar spacing (mm) p k = 0.3 mm p k = 0.4 mm p k = 0.3 mm p k = 0.4 mm 160 3? 40 300 300 200 ?S 3? ?S0 300 240 !6 ?0 ?00 ?S0 280 !? !6 !S0 ?00 320 !0 !? !00 !S0 360 S !0 S0 !00 Notes 1 ¹he 'norme|' ||m|t o¦ 03 mm mey |e re|exed to 04 mm ¦or `O end `C! exposure c|esses |¦ there |s no spec|¦|c requ|rement ¦or eppeerence 2 ¹e||e essumpt|ons |nc|ude c nom ?S mm end | ct,e¦¦ ( | ctm ) ?9 Mle This publication is available for purchase from www.concretecentre.com 205 Append|x C Les|gn e|ds C9 C9.1 C9.2 Design for axial load and bending General ln co|umns, des|gn moments / ld end des|gn epp||ed ex|e| ¦orce / ld shou|d |e der|ved ¦rom ene|ys|s, cons|deret|on o¦ |mper¦ect|ons end, where necessery, ?nd order e¦¦ects lt |s necessery to ce|cu|ete e¦¦ect|ve |engths |n order to determ|ne whether e co|umn |s s|ender (see lurocode ?, C| SS3? end lxpress|on (S!S)) ¹he e¦¦ect|ve |ength o¦ most co|umns w||| |e //?< / 0 < / (see lurocode ? l|gure S/¦) lL 66S/ |6| C| ?!0 suggests thet us|ng the procedure out||ned |n lurocode ? (SS3?(3) end SS3?(S)) |eeds to s|m||er e¦¦ect|ve |engths to those te|u|eted |n bS S!!0 |/| es reproduced |e|ow es ¹e||e C!6 lxper|ence suggests thet these te|u|eted ve|ues ere conservet|ve Table C16 Effective length e 0 : conservative factors for braced columns End condition at top End condition at bottom 1 2 3 1 0/S 0S0 090 2 0S0 0SS 09S 3 090 09S !00 Key Cond|t|on ! Co|umn connected mono||th|ce||y to |eems on eech s|de thet ere et |eest es deep es the overe|| depth o¦ the co|umn |n the p|ene cons|dered vhere the co|umn |s connected to e ¦oundet|on th|s shou|d |e des|gned to cerry moment |n order to set|s¦y th|s cond|t|on Cond|t|on ? Co|umn connected mono||th|ce||y to |eems on eech s|de thet ere she||ower then the overe|| depth o¦ the co|umn |n the p|ene cons|dered |y genere||y not |ess then he|¦ the co|umn depth Cond|t|on 3 Co|umn connected to mem|ers thet do not prov|de more then nom|ne| restre|nt to rotet|on Note ¹e||e te|en ¦rom /onao/ |o· |/e Je·/qn o| conc·e|e /a//J/nq ·|·ac|a·e· |o /a·ocoJe ? |3S| ¹he ve|ues ere those used |n bS S!!0 lert ! !99/ |/| ¦or |reced co|umns ¹hese ve|ues ere c|ose to those ve|ues thet wou|d |e der|ved |¦ the contr||ut|on ¦rom ed|ecent co|umns were |gnored Design by calculation Assum|ng two |eyers o¦ re|n¦orcement, / s! end / s? , the tote| eree o¦ stee| requ|red |n e co|umn, / s , mey |e ce|cu|eted es shown |e|ow lor ex|e| |oed N / sN /? (/ ld ÷ a cc n| c| /J c /g C )/(s sc ÷ s st ) where / sN tote| eree o¦ re|n¦orcement requ|red to res|st ex|e| |oed us|ng th|s method / sN / s! ÷ / s? end / s! / s? where / s! (/ s? ) eree o¦ re|n¦orcement |n |eyer ! (|eyer ?) / ld des|gn epp||ed ex|e| ¦orce a cc 0SS n ! ¦or < CS0/60 / |reedth o¦ sect|on Concise: Fig. 6.3 J c e¦¦ect|ve depth o¦ concrete |n compress|on l\ < / Concise: Fig. 6.4 This publication is available for purchase from www.concretecentre.com 206 C9.3 where l 0S ¦or < CS0/60 \ depth to neutre| ex|s / he|ght o¦ sect|on s sc , (s st ) stress |n compress|on (end tens|on) re|n¦orcement lor moment N / sM /? |/ ld ÷ a cc n| c| /J c (//? ÷ J c /?)/g C |/|(//? ÷ J ? )(s sc ÷ s st )| where / sM tote| eree o¦ re|n¦orcement requ|red to res|st moment us|ng th|s method / sM / s! ÷ / s? end / s! / s? vhere re|n¦orcement |s not concentreted |n the corners, e conservet|ve epproech |s to ce|cu|ete en e¦¦ect|ve ve|ue o¦ J ? es |||ustreted |n l|gures C4e) to e) So|ut|on |terete N \ such thet / sN / sM Rectangular column charts A|ternet|ve|y / s mey |e est|meted ¦rom co|umn cherts l|gures C4e) to C4e) g|ve non-d|mens|one| des|gn cherts ¦or symmetr|ce||y re|n¦orced rectengu|er co|umns where re|n¦orcement |s essumed to |e concentreted |n the corners ln these cherts a cc 0SS | c| < S0 Mle | y| < S00 Mle S|mp||¦|ed stress ||oc| essumed / s tote| eree o¦ re|n¦orcement requ|red (/ s | y| ///| c| )//| c| /| y| where (/ s | y| ///| c| ) |s der|ved ¦rom the eppropr|ete des|gn chert |nterpo|et|ng es necessery |etween cherts ¦or the ve|ue o¦ J ? // ¦or the sect|on / |reedth o¦ sect|on / he|ght o¦ sect|on vhere re|n¦orcement |s not concentreted |n the corners, e conservet|ve epproech |s to ce|cu|ete en e¦¦ect|ve ve|ue o¦ J ? es |||ustreted |n l|gures C4e) to e) J ? e¦¦ect|ve depth to stee| |n |eyer ? This publication is available for purchase from www.concretecentre.com 207 Append|x C Les|gn e|ds 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 K r = 0.2 N E d / b h f c k M Ed /bh 2 f ck 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 A s f y k / b h f c k h/2 h d 2 Centroid of bars in half section d 2 /h = 0.05 Figure C4a) Rectangular columns ] 2 /a = 0.05 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 K r = 1 K r = 0.2 N E d / b h f c k M Ed /bh 2 f ck 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 A s f y k / b h f c k h/2 h d 2 d 2 /h = 0.10 Centroid of bars in half section Figure C4b) Rectangular columns ] 2 /a = 0.10 This publication is available for purchase from www.concretecentre.com 208 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 K r = 1 K t = 0.2 N E d / b h f c k M Ed /bh 2 f ck 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 A s f y k / b h f c k h/2 h d 2 d 2 /h = 0.15 Centroid of bars in half section Figure C4c) Rectangular columns ] 2 /a = 0.15 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.1 1.3 0.3 0.4 0.5 0.6 0.7 0.8 0.9 K r = 1 K r = 0.2 N E d / b h f c k M Ed /bh 2 f ck 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 A s f y k / b h f c k h/2 h d 2 d 2 /h = 0.20 Centroid of bars in half section Figure C4d) Rectangular columns ] 2 /a = 0.20 This publication is available for purchase from www.concretecentre.com 209 Append|x C Les|gn e|ds C9.4 C9.5 C9.6 Biaxial bending in rectangular columns As e ¦|rst step, seperete des|gn |n eech pr|nc|pe| d|rect|on, d|sregerd|ng ||ex|e| |end|ng, mey |e underte|en No ¦urther chec| |s necessery |¦ 0S < l y /l . < ?0 end, ¦or rectengu|er sect|ons, 0? ~ (e y // eq )/(e . // eq ) or (e y // eq )/(e . // eq ) ~ S0 Otherw|se see Sect|on S63 o¦ Conc/·e /a·ocoJe ? lor squere co|umns (e y // eq )/(e . // eq ) / ldy // ld. Concise: 5.6.3 Circular column charts ln e s|m||er menner to C93, the eree o¦ re|n¦orcement ¦or c|rcu|er co|umns / s mey |e est|meted ¦rom the cherts |n l|gures CSe) to CSd) ln these cherts a cc 0SS | c| < S0 Mle | y| S00 Mle / s tote| eree o¦ re|n¦orcement requ|red (/ s | y| // ? | c| )/ ? | c| /| y| where (/ s | y| // ? | c| ) |s der|ved ¦rom the eppropr|ete des|gn chert |nterpo|et|ng es necessery J// e¦¦ect|ve depth/overe|| d|emeter Links l|n|s |n co|umns shou|d |e et |eest S mm or mex|mum d|emeter o¦ |ong|tud|ne| |ers/4 |n d|emeter end ed|ecent to |eems end s|e|s speced et the |eest o¦ !? t|mes the m|n|mum d|emeter o¦ the |ong|tud|ne| |er, N 60% o¦ the |esser d|mens|on o¦ the co|umn, or N ?40 mm N 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.05 0.10 0.15 0.20 0.25 0.30 K r = 1 K r = 0.2 N E d / b h f c k M Ed /bh 2 f ck 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 A s f y k / b h f c k h/2 h d 2 d 2 /h = 0.25 Centroid of bars in half section Figure C4e) Rectangular columns ] 2 /a = 0.25 This publication is available for purchase from www.concretecentre.com 210 K r =1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 1.0 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.2 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.9 d h Ratio d/h = 0.6 N E d / h 2 f c k M Ed /h 3 f ck A s f y k / h 2 f c k Figure C5a) Circular columns ]/a = 0.6 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.02 0.04 0.06 0.08 0.10 0.12 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0 0.1 0.2 0.3 d h Ratio d/h = 0.7 N E d / h 2 f c k M Ed /h 3 f ck K r =1 A s f y k / h 2 f c k A s f y k / h 2 f c k Figure C5b) Circular columns ]/a = 0.7 This publication is available for purchase from www.concretecentre.com 211 Append|x C Les|gn e|ds d h 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Ratio d/h = 0.8 K r =1 N E d / h 2 f c k M Ed /h 3 f ck A s f y k / h 2 f c k Figure C5c) Circular columns ]/a = 0.8 d h 0.05 0.10 0.15 0.20 0.25 0.30 0.7 0.2 0.3 0.4 0.5 0.6 0.8 0.9 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 Ratio d/h = 0.9 K r =1 N E d / h 2 f c k M Ed /h 3 f ck A s f y k / h 2 f c k Figure C5d) Circular columns ]/a = 0.9 This publication is available for purchase from www.concretecentre.com 212 Eurocode 2 resources Publications Concise Eurocode 2 CCll-00S, ¹he Concrete Centre, ?006 A hend|oo| ¦or the des|gn o¦ |n-s|tu concrete |u||d|ngs to lurocode ? end |ts 0l Net|one| Annex How to design concrete structures using Eurocode 2 CCll-004, ¹he Concrete Centre, ?006 Cu|dence ¦or the des|gn end dete|||ng o¦ e |roed renge o¦ concrete e|ements to lurocode ? Economic concrete frame elements to Eurocode 2 CCll-0?S, ¹he Concrete Centre, ?009 A se|ect|on o¦ re|n¦orced concrete ¦reme e|ements |n mu|t|-storey |u||d|ngs Precast Eurocode 2: Design manual CCll-0!4, br|t|sh lrecest Concrete lederet|on, ?00S A hend|oo| ¦or the des|gn o¦ precest concrete |u||d|ng structures to lurocode ? end |ts Net|one| Annex Precast Eurocode 2: Worked examples CCll-034, br|t|sh lrecest Concrete lederet|on, ?00S vor|ed exemp|es ¦or the des|gn o¦ precest concrete |u||d|ngs to lurocode ? end |ts Net|one| Annex Concrete buildings scheme design manual CCll-0S!, ¹he Concrete Centre ?009 A hend|oo| ¦or the lSructl chertered mem|ersh|p exem|net|on, |esed on lC? Properties of concrete for use in Eurocode 2 CCll-0?9, ¹he Concrete Centre, ?00S low to opt|m|.e the eng|neer|ng propert|es o¦ concrete |n des|gn to lurocode ? Standard method of detailing structural concrete lnst|tut|on o¦ Structure| lng|neers/ ¹he Concrete Soc|ety, ?006 A menue| ¦or |est prect|ce Manual for the design of concrete building structures to Eurocode 2 lnst|tut|on o¦ Structure| lng|neers, ?006 A menue| ¦or the des|gn o¦ concrete |u||d|ngs to lurocode ? end |ts Net|one| Annex BS EN 1992-1-1, Eurocode 2 – Part 1-1: Design of concrete structures – General rules and rules for buildings br|t|sh Stenderds lnst|tut|on, ?004 National Annex to Eurocode 2 – Part 1-1 br|t|sh Stenderds lnst|tut|on, ?00S Software RC spreadsheets: V3. User guide and CD CCll-00S ¹he Concrete Centre, ?006 lxce| spreedsheets ¦or des|gn to bS S!!0 end lurocode ? end |ts 0l Net|one| Annex Websites Eurocode 2 – www.eurocode2.info Eurocodes Expert – www.eurocodes.co.uk The Concrete Centre – www.concretecentre.com Institution of Structural Engineers – www.istructe.org This publication is available for purchase from www.concretecentre.com 3 Mombors of tho Stooring Croup ¦ohn Meson A|en bexter 8 Assoc|etes (Che|rmen) Stuert A|exender vSl Croup p|c le| Chene M|nere| lroducts Assoc|et|on ÷ Cement Cher|es Coodch||d ¹he Concrete Centre ¹ony ¦ones Arup Andy ly|e NkM Consu|tents k|cherd Moss lowe|| ¹o|ner Assoc|etes Nery Nereyenen C|er| Sm|th lertnersh|p k|cherd Sh|pmen LClC ko|ert Vo||um lmper|e| Co||ege, 0n|vers|ty o¦ london kuss vo|stenho|me vS At||ns 8 L¹l lro|ect O¦ucer kod ve|ster Concrete lnnovet|on end Les|gn Mombors of tho Concroto |ndustry £urocodo 2 Croup {C|£C} ¦ohn Moore Consu|tent (Che|rmen) C||ve budge br|t|sh lrecest Concrete lederet|on le| Chene M|nere| lroducts Assoc|et|on ÷ Cement ¦ohn C|er|e ¹he Concrete Soc|ety Co||n C|ever|y Construct Cher|es Coodch||d ¹he Concrete Centre le|g Cu|veness|en bkl Ceo¦¦ lerd|ng LClC ¹om lerr|son M|nere| lroducts Assoc|et|on ÷ Concrete ¹ony ¦ones Arup ¦ohn Meson A|en bexter 8 Assoc|etes k|cherd Moss lowe|| ¹o|ner Assoc|etes Nery Nereyenen C|er| Sm|th lertnersh|p k|cherd Sh|pmen LClC Mert|n Southcott Consu|tent kuss vo|stenho|me vS At||ns kod ve|ster Concrete lnnovet|on end Les|gn |nitia| soction drafts ! lntroduct|on Nery Nereyenen ? Ane|ys|s, ect|ons end |oed errengements Nery Nereyenen 3 S|e|s Cher|es Coodch||d 4 beems Cher|es Coodch||d, kod ve|ster S Co|umns ¹ony ¦ones, ¦ens ¹end|er 6 ve||s ¹ony ¦ones, ¦ens ¹end|er Append|x A Ler|vet|on o¦ ¦ormu|ee Cher|es Coodch||d, kod ve|ster, Owen broo|er Append|x b Serv|cee||||ty ||m|t stete Cher|es Coodch||d, Nery Nereyenen Append|x C Les|gn e|ds Cher|es Coodch||d, kod ve|ster This publication is available for purchase from www.concretecentre.com 4 workod £xamp|os to £urocodo 2: Vo|umo 1 1his pub|ication givos oxamp|os of tho dosign to £urocodo 2 of common roinforcod concroto o|omonts in roinforcod concroto framod bui|dings. v|th extens|ve c|euse re¦erenc|ng, reeders ere gu|ded through des|gn exemp|es to lurocode ? end other re|event lurocodes end re¦erences ¹he pu|||cet|on, wh|ch |nc|udes des|gn e|ds, e|ms to he|p des|gners w|th the trens|t|on to des|gn to lurocodes \o/ame ' wo·/eJ /\omo/e· |o /a·ocoJe ? |s pert o¦ e renge o¦ resources eve||e||e ¦rom ¹he Concrete Centre to ess|st eng|neers w|th des|gn to lurocodes lor more |n¦ormet|on v|s|t www.ourocodo2.info Char|os Coodchi|d |s pr|nc|pe| structure| eng|neer ¦or ¹he Concrete Centre where he promotes e¦uc|ent concrete des|gn end construct|on bes|des pro|ect meneg|ng end euthor|ng th|s pu|||cet|on he hes underte|en meny pro|ects to he|p w|th the |ntroduct|on o¦ lurocode ? to the 0l CCll-04! lu|||shed Lecem|er ?009 lSbN !-904S!S-S4-/ lr|ce Croup l © MlA ÷ ¹he Concrete Centre k|vers|de louse, 4 Meedows bus|ness ler|, Stet|on Approech, b|ec|weter, Cem|er|ey, Surrey C0!/ 9Ab ¹e| ÷44( (0)!?6 606S00 lex ÷44 (0)!?/6 606S0! wwwconcretecentrecom This publication is available for purchase from www.concretecentre.com