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# STRUC CONCRETE DESIGN CVEN 444

Texas A&M

GPA 3.53

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This 66 page Class Notes was uploaded by Carolyn Kuhn on Wednesday October 21, 2015. The Class Notes belongs to CVEN 444 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 43 views. For similar materials see /class/226127/cven-444-texas-a-m-university in Civil Engineering at Texas A&M University.

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Date Created: 10/21/15

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6 4 jg y 6 1 r 96 671065 M5 5amp5 rm M d 73425945 OEWU 67 2 78G6X 2453 L2 WFMI m MU Egg 4 55 x atix H DOLQJS LN 55555 7pc c c gt 30c 74 Ecx EsCJ39XS Es aimx 7 E T 74 lg 7 w C 54 w 715 M QM 9 o fag AM Ma 2 914441 2 amph 45 a A T d 70 T b a flow Awake W L3 at cum 1 M re flg i 1 k A quot3993quot 391 SCEYOZlv lfmi l 14573 iCl 4 Esq off 14d pjEs aLCA Sm ce 6 com Tag 1 quotCm1M6 Waf es AA Wm m gr M M 7 14 u a74c7 w a m L f ski Wlx 39 L 5 WM cracfw g MbMR Compression sude EGOA Neutral axis 39TA f 83 Tension side a b c d Figure 52 Stress and strain distribution across beam depth a beam crosssection b strains 0 actual stress block d assumed equivalent stress block mo093 e ozsfc i 0 44 E 1 lt c 1 4L 6 a L i 4 f2 T e Eggt 7 S fvm m ren Z c Z 39w ll mmr C T 085 100 M AND AS 10k0 mag hot 5 Mn nun2 an m w W5 A375 060 A 5 at Fig 48 quot1 Region shown in h Reinforcement i 39 P a Stage A Before cracking P I i i 1 z 41 1 1 4 1 i i i ii i i Xc39fL b Stage C After cracking before yield service load p l Crushed concrete p Shear cracks i T ZMM It w m p 39c Stage EFaiiure Figure 4 5 Cracks strains and stresses in test beam esgtey Strains Stresses Strains Compressive Tensile Strains Stresses 47 Stresses 4039 h E Failure D Reinforcement yields 30 g x c39 45 2 20 G Serwce load 5 0 E O E M M 10 y Axis of zero strain B Cracking HE A 0 l l l l l I I l 0 00004 00008 00012 00016 Curvature 1Iln Amwgke Bx tiC aAWf REQWREMERW 57 3 31111353257 2 743 h m s QQMWNTHEWWY 1 E43 ca a 2m 1amp5er wme n I I FIRM VJVlorvtngmnvr Anawarm 94 3amp4 2 823 39 39 Am ix2 cuO305 a a y T W m 16 39 H V P 6 a I at 44 T in I I q 2 41 a r R strep Iquot ASEUME T4 A13 z 37qu ng oooo 3m 1422 STEP 239 QWQ Q Egd UWENP D g 2 C T 93541 cm Q E 1299 we 2 a waf e m 435 amp mm 6151 3 Check 5 4 Wm WWMMQ I Assumamp M39Mtf rk Mushk3 3va m z 2 H m Wm stun quot C 555 A E 5w mm a m I iii 3 i MW I lmp Mh W C a 3939 gt mgzxziw g M MM 53 75 BMW TENSi M J Gammageiw g ngmagtz iiT N39Lum TYPES 33 FLEX UEEALr FAELDE E 1 TH TEA30 2 Keloms QM lO Plt 4Ve WCf 0W4 4W1 Mm raimCgrsgef gt 2 COMkO39TMiw z anemia drM mgrow g aak amici C ovbr mtva ha39rcrw V a I C a on mca z g 9 hawk CWR39quotLAAL Q 3 Sharkquot M Eum fc0003 H N I l M A l g gt0ooS 39 i 2 39 39 o T I quot s3 I I U c2 O O O O 0 i V I D 0 G d r r 3 whck39m 39quotSba bWAELE I O Q 936F3lt Qzawioaw 39qu 92535m wa 5 A r Mm uTn 9 0 Jilini vs Cc g okoqm Wagggfaqm AP IOI I W gr I W Na m w a Q l sliwhtil HGT grew33a 009455 N m 0 h m9 a W I m0 9 a R ma a WW mnF Q D me u IH nu K MOSHTWW gtnmv 097930Zm WDWWOF u Mum W4 Swag m2 A E1 gr um moon mlt H k 9000 dam 6F Er H 6amp3 40 WERV w P Adar rmr H a 0 m mufoow mm mdeOGIT EQUMM UM g CaMPAqngyurv Arr W HLUZE I C T 086 chla9 5 x Eggs C2 C16 atLC a uba z m m C74 Mtg 93 wquot9 tquot 53 33 1 z o6 c a EE Gmg 02 g cm 6116 4996 2 1 2 ES cu at jaw d 0 60 we er on J Lm at mwv M w 4 Mng av 23 T 740 T 74 Asa 5 094 am Jg mama now A Acr 31392 i 9 AOmm quot in magigx 1 WW 7 441 65 af rm i9ng r j g r g mak Egan f 903 MGL one oi HQ lollowt39nj g I i 65 gt 0005 7 Tmslm if M 5 01002 9 Marc 1 WMWQ 000143 85 539 OIOO S 7 Tina SIM so 0003 39 as 0003 c 0375d Tquot T I a c 06d 2 0005 at 0002 a b Figure 56 Strain Limits a Tension Controlled b Compression Con trolled 77577 SEWWK 9 ores 1 0451 335 g ow Spirt gm gt gm 5 0 1QT J Q39 4 057 67g 39 090 x I r e a n SPIRAL 070 048 835 065 OTHER Compression 4 T t Tension Controlled rans39 Ion Controlled 39 a 0002 8 0005 c c 0600 0375 d V d Interpolation on cdt Spiral 037 020c Other I 023 025cdt 4147ch 5441gg 69 039003 023H 0003 160 3 my Clo 4 1 2391on 81990 41 AF bo lax S k z 3 gang M ldw 14 39 085 73 FZ 40 5 ms aosC J ooo 6 4mm 4 gees iaab 0935 7 gt 86 psi Mihlmwm gm km wi C genial a 5 1 ampMfy mvi Eamp 39 ASImm 2 avg wi 2 EmaQua 747 7 Tw wm quot 9 quot 7mm quotgquot w j mm 2 g j 3 4i 2i AISlmr v x gt mg A Saigiagz quot a 53 a 31 W W Ex 2 mm git QM M MW E A i i r Qw aa igg f wgw gyma H May A ww nww Lam w Kama m M m 4 m 3 mm mm m a w A a A A A f MW n mfg x aw NQQIQ imwmg I g in la a Edi mi Kama W w i Q Ana 1555 5nsfgi1 gun me WM QM 9diAi galya E6 291990 4 392 R 3Wquot E Asymwx 3 I aj w 5 gaes oess msg ggjf 0 5 Evinkw e x 2 a 3 39 A quot 39 4 kg H11 571 i U A 541 Example 51 Fiexural Analysis of a Singly Reinforced Beam Tension Reinforcement Only A singly reinforced concrete beam has39the crosssection shown in Figure 59 Determine if the beam is tension or compressioncontrolled Given f 2 4000 psi 276 MPa determine if the beam satisfies the ACI Code if 39 a fy 60000 psi b f 40000 psi 39 60 p m 0033 From Eq 510a 3 V 4000 minimum allowable reinforcement ratio pm 3f fy 40 000 00047 Solution a fy 60000 psi Asfy 60 x 60000 a 085be 085 X 4000 X 10 10591n c 12461n c 1246 gt dt 180 069 060fromF1gure55 Hence As did not yield and the strain is smaller than 0002 inin Brittle behavior results as the section is compressioncontrolled and does not satisfy the ACI Code requirements for exurai beams 39 Solution b fy 40000 psi Bl lt c gtI 1 085f hi 539 E c 2 g i E s E 387 2 F m A st o 39 10 in quot7254 mm F e l a j lb 0 Figure 59 Stress and strain distribution in a typical singly reinforced rectan gular section a crosssection b st ains c stresses Photo 59 Crushing of concrete at compression side of beam subjected to exure Assume f5 fy 3V4000 00047 or I 200 0005 PM 40000 controls 9m 40000 39 X a w 706 in c 831 in 39 085 x 4000 x 10 e 0003 0003 M 00035 inin lt 0005 831 fy M 0 00137 inin lt 0 0035 EVE 29x106 39 39 c 831 046 gt 037 lt 060 d 180 5 Hence the beam is in the transition zone tension steel yielded But 6 lt0005 hence a re duced 4 for calculating Mu can be used Therefore ACI requirements for exure with 4 in the transition zone is satis ed However as previously discussed using a 4 value less than 090 exure is uneconomical Thus this section is uneconomical T 0 improve the design decrease AS or increase depth 542 Example 52 Nominal Resisting Moment in a Sineg Reinforced Beam For the beam cross section shown in Figure 510 calculate the nominal moment strength if fy is 60000 psi 4134 MPa and f is a 3000 psi 207 MPa b 5000 psi 345 MPa 0 9000 psi 621 MPa Solution b 10 in 2540 mm d 18 in 4572 mm A 4 in2 2580 m2 fy 60000 psi Note that fy should be in psi units in the pmin expression 3 fr 3 4 0 pm f 60000 0139 200 200 pmin fy 60000 00033 controls A 4 gt p bd 10X18 00222 00033 OK 31 085 Since the reinforcement is in one layer d d 18 in C 085fgba 085 X 3000 X 10 X a 25500a lb T Ajy 40 X 60000 240000 lb 240000 C T a 25500 1 39 a 94 11 39 c 31 085 1 m C 11 1 062 gt 060 compressioncontrolled d 180 Hence the beam is not ductile and does not satisfy the ACI 318 Code lt c gt gt 035 Hi T T a 2 SE c 339 L L 2 C 2 i i 1 j m NA 258cm2 da A4in2 N8 5 o o J i T l4 10in gtl 254mm a c Figure 510 Beam crosssection strain and stress diagrams Ex 52 a cross section b strains 0 stresses Solution b f 5000 psi 31 080 350 00 pmin 60000 00035 Actual p 0022 OK C 085 X 5000 x 100 42500 a 240000 T 240000 a 565 in 42500 a 565 c Bl 080 7061n c 706 gt lt dt 180 039 0375 060 Hence the beam is ductile but in the trasition zone with 4 less than 090 eey M Asfy d 40 x 60000 18 3642000 inlb 411 kN m Solution c f 9000 psi 31 065 3V9000 9quot 60000 00047 lt 0022 0k 40 x 60000 I a 314 In a 314 4 C B 065 83 1 d c e 0003 C 18 483 0003 W 00082 1nm gtgt 0005 gt 00075 hence tensioncontrolled namely ductile behavior 41gt 090 M 40 X 60000 18 3943200 inlb 446 kNm mm K g aw gmme w m 2435 in 3me mm mm n w n gnu Qw w 5 H a iv Ma imam u Emmimw a mi aw m N mm A N G 2 mth k g a M m w x E i Magma u in 3E 3 Eh i AW II l It 960 Unlv Law mun rm n I U mm Fl OOOuOiN 39 m Qaw WWW a N3 max Tilly H S 5 i mam wwwmvxw 5 a a a o m 50 n wwm awn Mum w 233 3 g w sw mm Em 333 N7 i Jug Cb Siam n c0179 quot152 o33 4k hf fh12 an o133 433c 2 33 0amp4 amp 2 A 4 7 1 4 1quot L C 0 0 6 0 Z y 51 8391 00020 SW51 TM Era 4 M3 As39wo x3 5020996400 at 6340 ka n 4 3 4 a M xol7 1 d 345 3924 MW CC MWMQ 20 0003 05751 tinMi E fe C MM 74 9 57 1 5m MW ne L 45 439 ML M77 08575159 9925 5 1224 Mn AW 0 1 5oy 2 gt 03 6303 144 H h Apt 6 0277 1 o a 57 0014 0130 I H 397 x10 I 04 m 60170 439 Sets 4000 3000quot 100 quot39 loco ESeiw a m s KlL Z Z3114 A q Mn 2 A20 7L MLL Prvcesg I la Fl a 2 am My Md 14442 mammt Madame C3 M W W dun 1Q MILL VomalMesz 9103 J b k A Accslyhak 4P0ukk5 are 1 H 5 A Weak 2 va eff bk 2quot 2 735 il 4 000 st I 1500 g CPIDW WWquot 40 433 a 0 Ad M 0 DWM JW NhP q 4amp3 H O39QS LLoLZECI39Aa z 2W 4 I W gigsfgbagz 3 Um w 1W0 E 44 a 0575 Jctu Wgzw39wW SF zi W gt Pig ag AW 39 WWW van quot Z bwk 5 Q M 5 Mg Mn M 8579 M m 45A zg gt W kFm 2 Mum 0 gt98 0 fi vct a 4 3 Leee 450 0375 is w39z ie M 74 74mg wameJ 7 M Ew gm f 7 C2 M 24 9 f ai 5 ma a beggiy g I mg 42 b 6935 7 a 4 54 3 quot 72 quot 795 ZHLyf 7 W gt 35ng 4 a 39 7 4L3 1Wquot Z Mn M W 3 1 V 2 14ng Asifm Azfmquot 74 591557454 QM Mn 39 w 7 ML 3 Tab1e 3 3 Hax1mum Number of Bars 1n S1ng1e Layer WILD I 39Mmd to stirrups a39Mmum main bars Minimum F 1 fr gt clear space L 1 3 Stirrup with 5 E 6 bars 4 Stirrup with V 7 11 bars 39ll db 39 L33 max 099 size Iquot for 3439I egg L33quot for quotogg 7 211 F19 32 1 Cover and Spac1ng Requ1rements for Tab1es 32 and 33 Ac quot119 Hax1mum s1ze coarse aggregate 34 in Bar Beam w1dth bw 1nches S1ze 1 10 12 14 16 18 20 22 24 26 28 30 5 4 5 6 1 e 10 11 12 13 15 1a 6 3 4 6 7 8 9 10 3911 12 14 15 7 3 4 5 6 7 8 9 10 1139 12 13 8 3 4 5 6 7 8 9 10 11 12 13 9 2 3 4 5 6 7 8 9 9 10 11 10 2 3 4 5 6 6 7 8 9 10 10 H1 2 3 3 4 5 5 6 7 8 8 9 Maximum size coarse aggregate 1 1nA Bar Beam w1dth bw 1nches S1ze 10 12 14 16 18 20 22 24 26 28 30 5 3 4 5 5 7 8 9 1o 11 12 13 6 3 4 5 6 7 8 9 9 10 11 12 7 2 3 4 5 6 7 8 9 10 10 11 8 2 3 4 5 6 7 7 8 9 10 11 9 2 3 4 5 6 7 7 8 9 9 10 10 2 3 4 5 6 6 7 7 8 9 10 11 2 3 3 4 5 5 g 6 7 8 8 9 bu 39 Ex Vat My efgw J4Lsc g sbkgg Mamas at 3 r 4139 4quot x it 4 g 3oamp 57 05401 72662 15 i 2 ggx z g 3 2265 p Saga 0 0J5 is 3921 0325 k Wk 2 2 0 60L 2 0325 55gt 3 339 kv Ma gag 2 y3a033 gtz 4 AWquot mn Z M 3 2 14h 9 A 5Q sag 154545 42 2mg 3 M4 0 5 cover gm 31 W 4255 In 0854X12351 kregr 97378 0393 15 43 AG Milklt3g 3 wwagt 43454 a msxz 9L3 1 364 M1 3 110 laws old 2 ASIP W 50427 32 m X 2337 111 439 5555 M 33 i WES 38 a23m b 0703 8 Z 3 l 2 24 5 05 Lg msm tg 1 if M 0123 0 21 139 707 JlI n 3 MW 454S In a w QJ AC 6ampwa far 390W Carob V ram MW CLEO MwnRum MW gmo A W4 W Qmomna 9161 Wm W W m M SUPWW WW5 I gIO 5 goum m 120 iM 12 Ufaa ZSAW CE Bad M o 2M5 al m 39aLM skim 3 a E 1 39 3 w I f I 1 1 BoJflx J S T CON M 00me le x Con WWW WW ComcaeTE CDVER AND R Zw py39u c 49 Reasomg par 00 1 x WW mcx yum mwt 5 39JDmto t Dr gszCC39WMt r64va s wagk k 12 93 AM i We Abrqgl h NEE st AUQ quotJP o ACE 7 7 Canquot fin 39 fame CW CAIfquot coWV 31 5 l o PAC MCa Le mbbub39o n3 AC1 j Q q Commit aw 6 exxgmw 9 M Maxwg l 130 14 5 4 c Commak my 24 2 W sabsmM439Jo aim or in what AUCL 0M 4 BaaM I 5 Nils 1 a 394 Fig 427 Example 44 2 No 8 bars 12 x 88 2dS 05 in st 05m 3No 9 bars 1 1 15 3 m 112 x 98 056 in 38 038 in 15 In I 15 in 15 in 98 in 98 in 38 in 38 in 019 in 019 in Not less man 1 ih nor db 98 in may hob 5 Sec 44 Design of Rectangular Beams 109 0 No Yes a Arrangement of bars in two layers ACI Sec 762 Larger of 133 diameter of coarse aggregate 332 1 in 762 l 32 T 1V in ACI 771 Larger of Bar diameter db 761 1 in 761 133 diameter of coarse aggregate 332 b Minimum bar spacing limits in ACl Code e qa ea Not more than the smaller of Fig 425 18 in Bar spacing limits in ACI 3 times slab thickness Code c Maximum spacing of flexural reinforcement in slabs 765 108 Chap 4 Flexure Basic Concepts Rectangular Beams 39Desiampr 31 Shaw Qai wce ed39aMVx r Bmm it 3mm Dung Mom MD LL Stead falls 70 1 56h i 930 4 404 UV ASSUWL to 119mi A V Ffm AsJ regp cq V L ASH gag 7 Asm w WW Meat Drcem tc VILLA Ffan QR me 139 Mpm Mrmv Z X m b mh k JzL Mama regkok Cover Rowui U Jro WM hwmlaor aNZQ Mawka d s M mm 2m no A t 5 met 10ng ici FIM M act Wm In r1 A5 Egg 4 A i 6 23 EOKZ w39gs39fc l MDgi ASJNA WCquot3157 L EL zE E LA 4 A 3 AS mCM 393 1 pmv 5 prm 39 ag iccl T4va FAS39G a 0L 139 239 R oL cover 0Q Su MA 924 391 x 3 Dz defS 1 I TF kg 0 otg nu mg k I 030 050 51 C i OW U113 M V Va Cv39u39 m 0 gt0 7c M DEQra Q ampt 4qu R39QJ VVGQOTCFQCR Pemsww BLOW I Awki A is baitdad or Qu s SxM 1 Dukrmlmx Mp Swat fc e Aesuvm b Compdte 4 10M k CM 1 1E I 931557 Mbcb QAGJQ 145 kayak 7 x45 mdm 45 R420 MeiT Qw kwcem 45mg pmv 5 375 M WIN 5 W V m dz 9L GoVuf quotdb LVW 0amp51C bdz haQ 8w Earn 160 35E 74MCvx 39139 V ipmd sPWquot af fc A re Myra a LaL ame 2 559ch 9 0quot 31560 4quot W 0 NM 1 2 33 4 5 lt6 m 400 K39Ft M WM Acr gamma Magma934049 49 050 Moi EC 7 039ampltg u ygt 22m 0 30 B QC 0003 4 402400 0175 baa MIXQ 400 Kz0 CIO a a85Fo w i 4 obS4 Yo75c 03 k 2 02552 Acf 619 M3 W 5 392 7 a 5265 Mdd 43 AC l E2 400 IQcg 40295XP xx1 aw 7 Z Zgt EltI4X2b9 quot63 a 3 2 1 gm 34075 jltl9msgt V79 39 s a D M 7 0 5 rs Q ee geeam A6 ref 55 92 v0 at 145an 7Com 553 m v 61 A L prov M ymv 97 1 0C a I Janet 09C5l5ggt4o 6ltl 0517942 40 I OIL 7quot N ngo m 41 4 lm Oh my l 7157 II t o o I yzu l ALI2n i 14 quot in quotA Jaws go Haws 0 4 O 860 I ll at 439 2L Ab 6 e 44wvfr m umueiyamezww 265 4 05 10 05 5 3 30 M k k NA m ewr g aaw Ma le 39 A 15 Lamar a 390 39 I I r MS KR 0 4 16w SUVCC M 395 Q so W Q u made r mac w M a MM kw 91424 2 L cg M Ook39 b H Cl at Jhr covek 9W NA Ab lzi39 24 L5 quot 0395 o0 05 2075quot 2 W gt h Olgsg c l o 1 3 mgvoiq 035C 4 HM go 552 ltOIEHTF 4i 4 03172 ab 0615 wk 1 A 45 Tr M44 3 Mt 39 2 a 400 OLD03 2 quot W 391 W31 M 4020X l Omil lt5gt 43 WV 3 CA5TltIO I39m a d 3 gwmr DOW 0 17 SLW 9 no gawk H 0pm he a M 9M 10 EM milks I 50 Jrv 5 5 5 7 HAumwthh A wow 00 lt 2 AMOPM wn k mch 3 Incrude 1Q 76 K3 im g W a 20 g mwc ataesng Hf I I Ill Malnsteel direction Slab thickness h Ar UnaquotW 5M IIIWQ 4 94457 2 6541114 5W a 561 Example 54 Design of a Oneway Slab for Flexure A one way singlespan reinforced concrete slab has a simple span of 10 ft 305 m and car 39 ties a live load of 140 psf 575 kPa and a dead load of 20 psf 096 kPa in addition to its self A weight Design the slab and the size and spacing of the reinforcement at midspan assuming a simple support moment Given f 4000 psi 275 MPa normal weight concrete fy 60000 psi 4134 MP3 Minimum thickness for deflection 2 Solution Minimum depth for de ection h 20 10 x 1220 6 in 1524 mm Assume for exure an effective depth d 5 in 127 mm 6 X 12 144 selfweight of a 12 in strip X 150 75 lbft 359 kNm Therefore factored external load wquot 1220 75 16 X 120 3381bft 338 x 102 factored external moment M 8 X 12 mlb 50700 in lb 57 Nm Mu 50700 required nominal moment strength Mquot 4 W 56334 1nlb Assume moment arm jd 090 d 09 x 50 45 in M Tjd Asfyjd or 56334 A5 X 60000 X 45 A 021 inz 12 in strip A 21 X 0000 xiv 9 6 031 in a 085be 085 x 4000 x 12 Recalculate A using the correct moment arm 56167 A X 60000 5 to give A 0193 in2 per 12in slab strip Use 4 bars at 12in center to center spacing Check strain 6 31 c aBl 8 5 037 in e 0003 0003 Lg 3 0038 inin gtgt 0005 inin hence section is tensioncontrolled cl 090 No 4 at 18 in C4 39 No 4 at 12 in ct main steel le 7quot 39ilb 3 a n n 4 T V 39 i punt 3 l No 4 at 12 in cc 1 A I 39 lt 10ft M m a l l v l l v39J LW No 4 at 18 in c4 temperature steel b Figure 515 Reinforcement details of the oneway slab in Ex 54 a sectional elevation b reinforcement plan Check Minimum Reinforcement 020 Actual p 5 0 x 12 00033 200 200 pm Ty W 00033 controls 3V 3V4000 00031 f 60000 39 Accept the design Shrinkage and temperature reinforcement p 00018 area of steel 00018 x 6 x 12 013 in2 No4 bars at 18 in c c Provide No 4 bars at 18 in center to center maximum allowable spacing 5h 5 x 6 30 in for temperature and shrinkage 7 Hence this design can be adopted with slab thickness h 6 in 1524 mm and effectivequot depth d 60 075 025 5 in 1270 mm to satisfy the i in minimum concrete cover rei quirement Use for main reinforcement No 4 bars at 12 in center to center and for tempera7 ture reinforcement No 4 bars at 18 in center to center as shown in Figure 515 Two Way Slab Example The general procedure for designing the two way slab 1 The limitations required by ACT Code are all met Determ1ne the minimum slab thickness us1ng the slab thickness equations Generally the slab thickness on a floor system is controlled by a corner panel as the calculation of hmin for an exterior panel given greater slab thickness than for an interior panel 2 Calculate factored loads wD wDL self weight of slab w 14wD 17wLL 3 The shear stresses in the slab are not critical The critical section is at a distance d from the face of the beam For a lft Width 1 l V w 7 7 beam w1dth ddepth of slab 2 2 compare with K viaEM determine Whether or not shear Will have an effect 4 Calculate the total static moments in the long and short directs 8 M W 11an 2 8 where 11 and 12 are length of the panels and lnl ll 7 c and lnz 12 7 c and c is the size of the column 5 Calculate the design moments for the long direction Distribution of the moments in one panel I Negative moment Mn factor Mol I Positive moment Mn factor Mol Distribution of panel moments in the transverse direction to the beam column and middle strips I Find the 1211 I Find ocl I Compute oc1lz ll gt 10 Distribution the negative moment Mn The portion 0 the interior negative moment to be resisted by the column strip is obtained from the AC1 tables by interpolation and is determined by I Column strip CS factor Mn I Middle strip MS factor Mn Since oc1lz ll gt 10 the AC1 Code section 1365 I Beam factor CS I Column strip factor CS I Middle strip MS Distribution of positive Mp The portion of the interior positive moment to resisted by column strip is obtained by interpolation and is determined by I Column strip CS factor Mp I Middle strip MS factor Mp Since oc1lz 11 gt 10 the AC1 Code section 1365 I Beam factor CS I Column strip factor CS I Middle strip MS 6 Calculate the design moment in the short direction same procedure as 5 7 The steel reinforcement requirement and number of bars in the long direction column and middle strip and in the short direction column and middle strip Find Mu Width of the strip b Effective depth depth of slab Calculate Ru Mu bd2 Find p Find A5 pbd Minimum A5 0002th Select the bars Determine the development length required for the bars shown q1ksi and fy 60ksi Check the anchorage in the column If it is not satisfactory design an anchorage using a 180 hook and check adequacy 2 in clear 15 in clear 10 bar 18 in h 39 L thehai ihehai 39 barplaeernent 13 pcoating 10 and 2 light weight concrete 10 and 10 bar is nlb 127 in lm 20 c 7 13101060000 7 7W7 1 6166db 21d 6166127 in 783 in gt Use 79 in 6166 fore there is spacing or 5 and The total length available in the column is l 51 in e 15 in 495 in there strength ofthe bars Howeverc1ltdb can be no greater than 25 So use 2 39 check and F10 bar size 11 3a y1fy db cK 40 V W a J 31310101060000 370 40J4000 25 1 370db 21d 370127 in 4698 in gtUse 47 in So you could argue that there is sufficient space if the transverse loading is enough to have cKLrdb gt 25 However a hook would be a better method to ensure that there is sufficient development length The development length is computed by zhd 1200 b J7 1897 4000 21m 1897db 2 lhd 1897127 in 241 in gt Use 25 in There is no at factor used in hooks and the other reduction factors are for cover yield strength ofthe steel fy 60 ksi 7 l concrete 7 1 and excessive reinforcement is unknown so use 7 l and ties assume 701 So there is sufficient length for a hook Considering the anchorage ofthe beam bars into a column determine the largest bar that can be used with out a hook fE 3ksi and fy 40ksi 2 in clear 1 ie T 18 in 15 in clear m m 39 39 u u u 39 A y 13 mowing 10 and A lightweight concrete 10 andld 1811715 in 155 in 17d a sz db zoJf 13101040000 3798 25 i a 3798 db 16 043 in gt Use 3 bar 0375111 3798 A simple supported uniformly loaded beam carries a total factored design load of 48 kft including selfweight on a clear span of 34 ft fC 3 ksi and fy40 ksi Assume that the supports are 12 in wide and assume that the bars are available in 30 ft lengths Design a rectangular beam Determine bar cutoffs Locate splices and determine the lap length The load for a simply supported beam and design for the positive moment of the beam and 48 ldft wuzn2 48 1dft34ft2 8 8 6936 kft M Assume that the tension controlled so that I 09 Assume a k025 and 31 085 because fc 3 ksi so that k 31k 0850250 02125 R 085fck 1 0853 ksi021251 03922i5 04843 ksi You can use various k value to design the beam as long as k lt 0375 Find bd2 from b 055 d u M M U bd bd R vi R 6936 k 12 in bdz 09 1 ft 04843 ksi 19096 in3 Assume that b 055 d and solve for d 055613 19096 in3 39 3 d aim 3262 m 055 b 0553294 in 1794 in and Use d 325 in and b 18in The dimensions and maximum moment have been determined Assume that k 02125 and a k d 02l25325 in 6906 in and compute the amount of steel needed 12A085 ba34w Y A 0853 ksi18 in6906 in s 40 ksi 792 inz Select 8 9 bars so that AS 8l00 inz 800 inz Compute the new a and c values fyAs 40 ksi800 inz 085be 0853 ksi18 in 6972 in a 6972 in c 8202 in A 085 Check the pm and the beam will meet the condition ACI 1051 200 0005 fy 40000 pmin larger of 3 pmin 0005 3 4 000411 fy 60000 Check the p value A 800 in2 p E 18 in325 in 0013675 2 0005 OK Check to see if the steel is in the tension controlled 3 0252 g 0375 ok d 325 in 01 gtd CJOIOO32325 ni9021n0003 c 1n 000887 2 0005 OK Compute the Mn for the beam M fyA oi g 40 ksi800 in2325 in 92845 kin 6972 in 2 The Mu for the beam and error is Mn Mn 0992845 kin 83561 kin 2 6963 kft J 6963 kft 6936 kft 100 039 6936 k ft The beam is overdesigned by 039 which is less than 10 Check to see ifthe bars will t in the beam using 4 bars 9 with 4 stirrups per row S b ndb 2cover ds mp n 1 18 in 41128 in 215 in05 in 3 316 in and spacing will be 316 in between bars So you could have bars next to one another for splicing For both cutoff and splicing on needs to know the development length of the bars Determine the development length for the barsThe bar coefficients are 01 bar placement 10 coating 10 and 7 light weight concrete 10 and 9 bar is db 1128 in l ny al b 20E 10101040000 20m 21d 3651db 21d 36511128 in 4119 in gt Use 42 in 3651 Determine the capacity of 1 bar a Mum Msfy 61 5 09100 in240 ksi325 in 10445 k in 3 8704 kft So the Mn 1 bar 8704kft Mn 2 bars 1741 kft Mn 3 bars 2612 k ft and Mn 4 bars 3482 k ft Mn 5 bar 4352kft Mn 6 bars 5224 kft Mn 7 bars 6093 kft and MH 8 bars 6963 kft Moment Diagram 8 bars 6963 k4quot 5224 k4quot Mnmenl HI 3483 k4quot U24EElEIlZlAlElEZDZZZAZEZESDSZSA anzlinn my Determine the development length and la for the bars The extension length 12 is either 12db121127 in 1352 in or d 325 in Use 12 325 in or 33 in Determine Where the bar cutoff points meet the moment diagram Obtain the equation for the moment M x816 kx748 mtg Set the moment equal to 5524 k and 3483 k 2 5224 k 816 kx748 kn x2 734x 21767 0 2855 R or 10265 in 2102 in 2 2545 h or 3054 in 2 306 in 2 3483 k 816 kx748 k x2 r34x145125 0 2500 h or 60 in 260 in 2 290 h or 348 in 2348 in The location ofthe cut off for a single bar is 102 in 33 in 69 in 5ft 9 in and 306 in 33 in 339 in 2811 3 in With a bar length of 22 h The location for the bar to be illy developed is 69 in 42 in 111 in and 339 in 742 in 297 in Moment Diagram 8 bays 6963 k4quot E has 5224 k4quot 4 bays Mnmenl HI 3483 k4quot u 2 A E a 11214mmznzzuzezaanazm anzlinn lj The location ofthe cut offfor a4 bar is 60 in 33 in 27 in 2 3 in and 348 in 33 in 381 in 3111 9 in With a bar length of 295 h The location for the bar to be illy developed is 27 in 42 in 69 in and 381 in 742 in 33939 Moment Diagram Mnmenl HI E24EElEIlZlJlElEZUZZZAZEZESUSZSA Lmunnm For lap splice to be a Class A splice in tension the area ratio must be twice be twice that of the required area The cutoff of the 2 bars from 6 to 4 bars have a total length of 295 ft so no splice on the bars were need The remaining 4 bars can be 30 ft long with an overlap ofld or 42 in there is a space of 24 in on either side ofthe bar for the full 34 ft so the lengths ofthe added bars are 24 in 42 in 66 in with splice length 42 in Given a simply supported beam with a load wd 18 ldft including selfweight and a live load WL 24 ldft fc 4 ksi fy 60 ksi d 215 in b 12 in AS 508 inz and In 24 ft wu 12wd 16wl 1218 ldft1624 ld 6 ldft Compute the moment and shear forces wulnz 6 ld 24 ft2 M 432 k 8 8 mzquln6ld 24 72k The equations for shear and moment are V x 72 k 6 1dftx 2 M x 72 kx 6 10ft The shear and moment d from the wall are Vux72k 6ldft2151n If D6L25k 121n 1ft M x 72 k215in1 6 WWW 11937 k in Sim li ed Method Compute the Vc of concrete simple method Vc 22Jfbd 210J4000 12 in215 in 32634 1b 2 326 k Compute Vn for the problem Vn M8L67k 075 Vn is greater than 05 VC so that shear reinforcement is needed Vs Vn Vc 8167 k 326 k 491 k You can use a ratio for stirrups to nd d SzAvfy 34 Vs 491k VS 3 fyd60 ksi215 in 39 Z 00381 i 1n The minimum value i M 001in 2 lt 00381in 2 which will control the amount of steel 3 fy 60000 in in Use a 4 bar Ab 02 inz so Av 202 inz 04 in2 Avfyd 04 in260 ksi215 in S T 491 k s 1051 in Check maximum V5 V5 491 k s 4J7cbd 4J4000 12 in215 inm 653 k Therefore smax im1075 in smax min 2 2 3 smax 1075 in 24in You can use 4 105 in however you would probably use 4 9 in Re ned Method Compute VC Vc 19def 2500pw 3quot bd s 35de A 39 2 pw s k 10197 2 bd 12 in215 in Vud 6125 k215 in Mu 11937 k Vc 19m25000019709212 in215 in s 3512 in215 mm 4271 ks 5711 k Vn is greater than 05 VC so that shear reinforcement is needed Vs Vn Vc 8167 k 4271k3896 k You can use a ratio for stirrups to nd M 3 AV Vs 3896 k V s fyd60 ksi215 in 092 S 10 2 003021L 5 in The minimum value 50 12 39 2 39 2 i A 0011i lt 003021i which will control the amount of steel 3 fy 60000 in Use a 4 bar Ab 011 ml so Av 2011 inz 022 in2 Avfyd 022 in260 ksi215 in s V 3896 k 728 in s Check maximum V5 V5 3896 k s 4 bd 4J4000 12 in215 in Therefore smax 65 k 10001b im1075 in ax mm 2 2 3 smax 1075 in 24 in Sm Use 3 7 in spacing you would probably use 3 6 in For variable spacing of the stirrups you would compute VS at various locations You would start out nding the equation for the shear envelope From using a pattern loading for the live load covering half the beam the maximum shear would be 16 24 ldft24 ft8 1152 k 72k 1152kx 0ft 12ft 72 k 5041dft x V x 72k 5041dft x Km 5 075 H 96 k 672 1dftx Vux72k Use 3 stirrups for the beam at d 215 in 1792 ft Vn 8396 k and AV 022 inz The rst shear value has spacing of 4 in from the support spacing in AV inz VS kips X ft 3 1 4 022 7095 0 3 12 5 022 5676 4583333 3 11 8 022 35475 1125 3 2 9 022 3153333 12 The 1St column is the bar number the 2quotd column is the number of bars the 3rd column is the spacing the 4Lh column is the area of the bars the 5Lh column is the VS generated by the bars and the 63911 column is the location of the end of the bars 35475 k VS 2 Avfyd Z 022 in260 ksi215 in S 8 in The plot ofthe shear design is Shear kip 100 90 80 39 70 i 60 50 4O 3O 20 Shear Designed shear reinforcements Location ft

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