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PROJECT NAME: HULTS RESIDENCE <br /> JOB: X10-2854 <br /> Required footing width base on 1500 psf; <br /> b = (6703 / 1500)1/2 = 2.11 ft or 2'-2" <br /> Reinforced Concrete Footing Design <br /> Footing - Design <br /> Service loads <br /> P= 6703 lb <br /> M= 0 ft-lb <br /> Factored loads <br /> Pu= 10725 lb <br /> Mu=oft-lb <br /> Compute width of ftg <br /> pa= 1500.00 psf <br /> H=8.00 ft <br /> tf= 12.00 in <br /> Pnet=Pa- Witg- Ws <br /> Pnet= 1500 psf- 12.00 in./12(150 pcf)-7.00 ft.(0 pcf) = 1350 psf <br /> eccentricity, e= M/P= 0/6703 = 0.00 ft <br /> pnet= P1(BL )±6Pel(BL2) <br /> p-min = 1385 psf <br /> p-max= 1385 psf <br /> p-max= 1385 psf< pall = 1500 psf FOOTING SIZE OK <br /> Check, 2.20 ft x 2.20 ft x 12.00 in <br /> Compute factored pressures <br /> p„= PI(BL )±6Pel(BL2) <br /> Pu= 10725 lbs <br /> Mu=oft-Ibs <br /> Max,p = 10725/[(2.20)(2.20)] + 6(0.00)(10725)/ [(2.20)(220)2] =2216 psf <br /> Min,p„= 10725/[(2.20)(2.20)]-6(0.00)(10725)/[(2.20)(2.20)2] = 2216 psf <br /> Check one-way shear <br /> x= minIL/ Ip„min +p maxi = 1.10 ft. <br /> Estimate, d <br /> CUSTOM DESIGN & ENGINEERING, INC. <br /> October 20, 2021 Page 10 / 17 <br /> Copyright©2021 Custom Design&Engineering,Inc.All rights reserved. <br />