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ARCH3355_2013_04_16_Foundation_Design

two way shear soil design pressure is calculated

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Unformatted text preview: ure is calculated over the area outside the square [efgh] -  Bottom reinforcement is usually required to resist the moment forces. compression in top of footing d d column load [P] b distance “d” from the face of the wall [critical section for shear] a d/2 e g face of wall [critical section for moment] f h Determination of the Footing Thickness: With no transverse reinforcement, the required thickness is determined by the tension stress limit of the concrete in either flexural stress or diagonal stress due to shear. Typically only providing transverse reinforcement if the cantilever edge distance exceeds the wall thickness. With transverse reinforcement is used, the critical concerns become for the shear in the concrete and the tension in the reinforcement. ACI code recommends limits of 8in for unreinforced footings and 10in for reinforced footings. d one-way shear line c upward soil pressure Selection of Reinforcement: Transverse [flexural tension and development length] and longitudinal [providing minimum for shrinkage] reinforcement. Minimum coverage is 2in from formed edges, 3in for unformed edges [footing bottom]. NOTE: Transverse reinforcement development length of should be coordinated with any doweling that needs to happen. 3” cover two-way or punching shear line effective depth “d” is the distance from top of the footing to the centroid of the reinforcing steel ld d tension reinforcing face of column: [critical section for moment] reinforcing steel tension in bottom of footing compression in top of footing compression in ARCH3355 | Construction III | Spring 2013 FOUNDATION DESIGN | P. RAAB ARCH3355 | Construction III | Spring 2013 12” 12” top of footing d 12” column load [P] b FOUNDATION DESIGN | P. RAAB d a b d/2 36” 5’ below grade d e #8 bar [1in dia] d f g Designing Wall Footings: Example tension reinforcing d face of column: [critical upward soil pressure section for moment] f’c = 3000psi, 12” fy = 60,000psi, 14,000 lb/ft of wall length, 7,000 lb/ft of wall length, #4 lb/ft2 3,500 bars [7in o.c.] 12” 36” d d d d 14” Step 01: Determine I...
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