Lab2sol - UNIVERSITY OF CALIFORNIA, BERKELEY Department of...

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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 Material (SEMM) Spring Semester, 2009 CEE 123 Laboratory Project No. 2 (Solution) Part 1: Dead Loads a) Floor Area Tributary to Each Member in sq. ft. Slab (1' wide strip): T A s = 1ft*1ft = 1ft 2 Joist (2' spacing * 20' length): T A j = 2ft*20ft = 40ft 2 Girder (20' between center lines * 20' length): TA g = 20ft*20ft = 400ft 2 Column (20' between column center lines, in both directions): T A c = 20ft*20ft = 400ft 2 Effect of Distributed Dead Loads Additional dead loads Floor finishes FF = 6lb/ft 2 Ceiling finishes CF = 11lb/ft 2 Fire sprinkler system FSS = 6lb/ft 2 HVAC HVAC = 12lb/ft 2 Partitions P = 20lb/ft 2 Total additional dead loads: DL add = 55lb/ft 2
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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 Material (SEMM) Spring Semester, 2009 1) Slab Distributed dead load acting on slab DL add/s = 55lb/ft/1’width = 55lb/ft 2 2) Joist Tributary width of joist TW j = 2ft Distributed dead load acting on joist DL add/j = DL add *TW j = 55lb/ft 2 *2ft = 110lb/ft 3) Girder Tributary width of girder TW g = 20ft Distributed dead load acting on girder DL add/g = DL add *TW g = 55lb/ft 2 *20ft = 1100lb/ft 4) Column Distributed dead load acting on column DL add/c =DL add *TA c = 55lb/ft 2 *400ft=22000lb b) Concrete Self Weight Supported by Members 1) Slab
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This note was uploaded on 02/27/2011 for the course CE 123 taught by Professor Mosalam during the Spring '09 term at University of California, Berkeley.

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Lab2sol - UNIVERSITY OF CALIFORNIA, BERKELEY Department of...

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