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by: Marianne Marvin III


Marianne Marvin III
GPA 3.53

A. Okeil

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About this Document

A. Okeil
Class Notes
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Popular in Civil Engineering

This 2 page Class Notes was uploaded by Marianne Marvin III on Tuesday October 13, 2015. The Class Notes belongs to CE 4410 at Louisiana State University taught by A. Okeil in Fall. Since its upload, it has received 25 views. For similar materials see /class/222646/ce-4410-louisiana-state-university in Civil Engineering at Louisiana State University.




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Date Created: 10/13/15
E 57000 f c 33w15 fc fr 75 f c fsp 67 ftquot M3 3 I Adz I T if section lies on NA M rrltlggtn5 3th E M M I 7 3117 7 311 Mam Ma 7 Stresses f 77quotkd fr 75fc39 nMn fs I d 7 ka Allowable Stresses ffrazz 045f5 7 20 ksi forfy S 60 ksi fs39 7 24 ksi forfy gt 60 ksi Service Moments ifc39all In MC 7 kd I MS fsrall 011 a 7ka Crack Widths 40000 40000 5mm15 f 725Ccor12 f 2 f5 gfyl Skin d effective depth Crack Widths w 0076 szdC A 103 12 for beams 135 for slabs dc dist from surface to closest steel f5 06fy A 7 effective tension area 7 of bars Deflections A 5 wL x 3 84 E1 1 wL my 1 wL my A dile 9quot 7E A d71PL3 9quot i Moments Mmax 7PL 7wy Mmax 7WU Mmax M iPL max7 4 Beams Normal analysis Assume E gt 5y 6 T 085fc ab Asf a 1cwhere 085iffc39 41m 85 7 fc 7 4005if4 lt fc39 lt 8 65iffc39 gt 8ksi 6 in 7 7 7 Twhereem 7 003 If E lt 5 restart assuming 6 lt 5 fs EJES9Use fs forfy Es 29000 a CU Q a TW w compression steel assume 5 gt 5y 5 gt 5y n C Cs T Cs Aslfy but ifs ltEy recalculate cwith E 003ampfsl EESamp a 15 Mn Ccd7Csd7dl Tmax Asm Tmax Cmax fy EI 0003 7 0003 d 7 c E 0004 for Asmax to find c a and Cmax 200 W020i 3 fc39 TauNi Asmm max DESIGN Beams Find As given dimensions u 14Du 12D 16Lu 12D 16W 1L DMn Mu a 09 if e gt 0005 a 048835if 0002ltelt0005 7 z 7 Mqu Rectangular9Mn 7 pmafybd 1 17M Pbd Am b7Z1SZ 3 r d Sm 11 of bars 7 1H 5mm 7 max db 1 Layer 7 15 Z db 2 Layers n115i 1dbnz15i15db1 Y a z a quotm 39 n1 in 1 Z layer nZ in 2 layer 710271 a z h 7V9Then normal analysis to find Mn9 Mn gt Mu careful may have to redo 11 Given Mu design As b and d 39 003 pde 85 1 EXMDS E 007 faIlure Mu z i mfy 2 a pd fybd 1 17fcsolve for bd 1 5 lt h lt 2 39 b pde bd Ameq spacing dact checkthrough normal analysis Given Mu with Fixed dimensions 785 fr 003 PMquot A f 0005003 Pee fy Z 7 Mn meybd 1 WC ileMn lt Mu9singe reinforcement not enough AMiMu lDMnicoi dC 7 AM F q 5 quot S ded39 i 005003 check strains A 7 Cs 5 7f5 Cs AsAslAsZ pbd fy check spacing on bottom dim h 9 7 Lila 15 38 05d band finish with check Slabs l hmm round up Self weight9139 1 h 150 l17W Live oad 9AreaLL LL 9 Factored loads u9get Wu 2 9Design for max of Mums will M1407 Wulnz 1e ln clear span length 9Mn gt if assumesI 007faiure f 003 pde 85 1f WHOM Gusually 01455 Mn pd fybdz 1 712 12Hwidth ofsec and solve for d hsq d 1 9 if Hgqlthmm use hmm Mn pequbdz 1 712 12Hwidth of sec solve forpma 200 7 l fy pm1 max ifgtpreqcontrolsfor posampneg As pb h 7 1 Ba rsft AsAlm Spacing 12 inchesBarsft round down Transverse Reinforcement AM 0001812 h Barsft AshAlm Spacing 12 inchesBarsft round down TSh bw 8mm 8mm 1 1 f Be f min 1quot 2 571th Z Szm L h 4 Ycrr or Z and take an average 9d pick bars get Sm check 5mm get dim 2x3 7 15 a 1quot normal analysis to get c a and checking strains f l a l falls in flange tbeam acts like rectangular beam bw be and lDMn lDAsfy d 7 If given crazy tshape j Mn Tmoment arm moment arm d 7 j Shear nominal shear stress 17 2 fr39 total conL39rete resistance Vc Zwad VnVcVs Vs Factor oads9getVu9CDVn 2 Vu Vn 2 vi Vc z f Vu no shear resistance needed Vc Mf Z reinforcement needed lt Vu ltPVL minimum shear f Vu gt ltPVL special shear reinforcement needed Vu S lDVn 4gtVc Vs Vs g7 V D 75to account for brittleness of shearfailure s f was Min reinforcement getAvmm max y 75Jf739bws fy 9 Pick bars amp f of legs set to Avmm amp solve for S 24 9Check5m min i 2 Special reinforcement9get Vs9checkAC limits 8fc bwd 0K 4fc bwd 0K 24 a 2 If Vslt 8 fc39bwd 0K Vs gt 4fc bwd NOT 0K Vs 9allows relaxed spacing simin F A d s i A legs Am 5 12quot 95mm min 2 4 8 b d Ist f W NOTOK9Beamisrejected 4 1 6de Critical section for shear 9factor oads9find Vum 9find critical section Xcr hwy d9VuT Vuym r WuXL39r9Then design for shear Bond RULE 1 3 Ivzlvem f E J W Developement length ld db 5 130 forreinfm above gt 12quot of conL39rete 100 otherwise other coated bars 15 apoxywcover lt 3db orS lt 6db lye l uncoated 12 10 q 7 8 for 6 bars ampsmaller 3 7 10 for 7 bars amp larger 1 13 for light weight conL39rete 1 0 for normal weight concrete 6 smaller of bar spacing or cover dimension A f kn W s maxsstirmp 71 Zir rup k 5 lt db 7 25 C kn 15when given no info on stirrups 9geth8ltMumu 9lDMn Z Mum 9get pm 9get152 911 quotquot9quot 11 Amp RULE 2 bars must extend 12db 7mmquot d PPWM 7 W 9get lDMn9 Mn ZM9getx9sti add max 1227 gtmax distm x max 12219use greater of rue1 or rue2 RULE 3at least As should extend into supports Columns Tied sszfc jug A fyAPnom 8Po 9Get PulDPnomax Z PuAA7 bh lt1gt885fc Ag 7 A fyA 2 Pu 9min dimensions of column 12quotX12quot 9min barsr 44 bars


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