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Lab_8_sol

# Lab_8_sol - UNIVERSITY OF CALIFORNIA BERKELEY...

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UNIVERSITY OF CALIFORNIA, BERKELEY Instructor:Khalid M. Mosalam Department of Civil and Environmental Engineering CE123: Design of RC Structures Structural Engineering, Mechanics and Materials (SEMM) Spring Semester, 2009 CEE 123 Laboratory Project No. 8 (Solution) Problem 5.2 f '’ c =4000psi f y =60,000psi w u =1.2*0.72+1.6*1.08=2.59Kips/ft M u =2.59*25.75 2 /8=215Kip-ft b=13in h=19in Æ d=16in M u = Ф bd 2 R Æ R=2.15*12000/(0.9*13*16 2 )=861 Æ ߩ and required A s =3.52in 2 and provided A s =3.81in 2 ൌ 0.0169 ଶ.ହସ 2 No 10(32) A s =2#10=2.54in 2 Æ ߩ ൌ ଵଷכଵ଺ ൌ 0.0122 ݈ ൑ 1.3 180 ൈ 12 33.3 a) R=650 Æ M u =650*0.9*13*16 2 /12000=162Kip-ft 162/215=0.75 So, one No10 bar can be cut at 0.25*25.75=6.44ft from the support Centerline ADD: d=1.33ft or 12d b =12*1.27/12=1.27ft Actual cut point=6.44-1.33=5.10ft from support Centerline b) For No10 bars, l d =60in (Table A.10) For No10 cut bar, provided l d =60(3.52/3.81)=55in=4.6ft Have (25.75-2*5.10)/2=7.78ft>4.6ft OK For No10 continued bars we have 6.44*12+4=81in>60in OK c) M u =162/0.9=180Kip.ft V u =2.59*25.75/2=33.33Kips ൅ 4 ൌ 88݅݊ ׎ܸ ൌ 0.75 ൈ 2√4000 L d =60in Æ Requirement satisfied d) V u at theoretical cut off point=(2.59*25.75/2)-(2.59*6.44)=16.7Kips

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