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<br />I <br />j <br />.1 <br />I <br />'j <br />I <br />! <br />I <br /> <br />ANALYSIS OF FOUNDATION <br /> <br />, .~ <br /> <br />Stemwall is 8 inch concrete masonry, filled with grout, 16 inches <br />high" <br />Footing is 16 inches wide by 9 inches deep <br />Earth;~over over top. of footing is 10 inches <br />I <br /> <br />Total:1 uniform wind uplift on foundation =~~ounds per linear <br />f 00 t. . <br />Uniform dead loads in pounds per linear foot: <br /> <br />. <br />. <br />. ~ " <br /> <br />i <br /> <br />3'7.. <br />WALL = ' ~l <-)' <br />F~;;~:----;;~~---------- <br /> <br /> <br />, . <br /> <br />l <br /> <br />ROOF = <br /> <br />, <br /> <br />" <br /> <br />: _0 <br /> <br />Total uniform dead load times 2/3 ==2~~pounds per <br />Net uplift force at top of foundation =~pounds per <br />Weigh~.'of stemwall footing earth x 2/3==?~pounds.per <br />Net upli ft at. bottom of footing == (:) pounds per <br /> <br />linear <br />linea r <br />linear <br />linea r <br /> <br />ft. <br />ft. <br />ft. <br />ft. <br /> <br />'-' <br />" <br /> <br />SUMMARY OF ANALYSIS <br />FOUNDATION IS STABLE. <br /> <br />" <br />,i <br />'.; <br />1 <br />, <br />i <br />, <br />1 <br />1 <br />.I <br />I <br />I <br />1 <br />} <br />.' <br />~ <br />( <br />. <br /> <br />. <br /> <br />ANALYSIS OF REINFORCING STEEL <br /> <br />Total uniform wind uplift on foundation == L"?(gounds per linear <br />foot <br />Uniform dead loads in pounds per linear foot: <br /> <br />ROOF == _ B'L pi f <br />WALL =~ plf <br /> <br />TOTAL= "'J<-r7 pI f <br /> <br />" <br />.' <br />'. <br />.' <br />'f'; <br />;~ <br />, " <br />'11 <br />i! <br />i <br />.1 <br />" ~~,' <br />, . <br />t, : <br />, <br />:j: <br />~,'t. <br />, ~: <br />u: <br />~ <br />'I <br />ij <br /> <br />Total uniform dead load times 2/3 <br />Net uplift force on foundation <br />Weight of concrete block stem wall <br />times 2/3 <br />Net uplift at t,op of footing <br />Total uplift forceon'each re-bar <br />Safe tension value of each re-bar <br /> <br />==1Io~pound5 per linear ft. <br />= 0 Pounds per linear ft. <br /> <br />= 5 J pounds per 1 i nea r ft. <br />== 0 pounds per linear ft. <br />=~ounds <br />=~pounds(incr.by 1/3). <br /> <br />SUMMARY OF ANALYSIS <br />REINFORCING STEEL SATISFIES ALL CODE REQUIREMENTS. <br />251( (/Y1 U-'. L~ ~ <br /> <br />" <br /> <br />1'- <br /> <br />----'.'...,----,.--.. -.....-,..... ....- .,. <br /> <br />ii',"" L <br /> <br />................ -.r <br /> <br />""-.-1'-"'. <br /> <br />_.._-~~--"-_..,._.,----_.._----_.__.._..,-~ -----~--,.- ---- <br /> <br />._.....-.-......._--~...."..,.-.,~."~----_._..... <br /> <br />~ ~. <br /> <br />---.-------,-__.~__.n <br />