GRAIlvT PRESSURES IIST STORAGE BUSTS. 13 



From Table 3a: A wall 6 inclies thick with a depth to steel of 5 



inches has a resisting moment of 32,300 inch-pounds, requiring 0.462 



square inch of steel per foot in height, which is sHghtly greater than 



actually required. 



LxP 

 Usiag the formula a = — o~ 



. 300X10X10 ^ ,, . , 



we have, a = 'jacoo — = 0.43 square mch. 



given by f " rounds 8 J" on center. 



If it is desirable to have the bars spaced in even inches, we can 

 increase the spacing to 9 inches and investigate to determine the 

 corresponding stresses in steel and concrete. 



The area of f " rounds 9" on center equals 0.41 square inch. 



0.41 

 p = -^^ = 0.0068. 



for p = 0.0068 and n=15, fc = 0.361, and y = 0.879. 



r M 30000 _„,^ , 



-^-=^^^^7^^0:4 1 X 0.879 X 5 =^^^^^Q P^^^^- 



r 2x/sX29 2X16640X0.0068 ^^^ . 



/c = j^ = ^-^ — — = 630 pounds. 



The stress in the concrete is less than the allowable assumed and 

 the stress in the steel is not excessive. 



To find the height at which the spacing may be increased to 10 

 inches: The area of f'' roimds 10" on center equals 0.37 square 

 inches. 



Using the formula L = — j^ — 



we have i = °-^I^^ = 0.37 X 700 = 259 



ts = -t^ = 25.9, for which 7^ = 2 (see Table 1). 



H, then, is equal to 20, and at a distance of 20 feet from the top we 



can increase the spacing of the bars to 10 inches. 



5'' 

 The area of - rounds 12" on center equals 0.31 square inch. 



8 



Z = 0.31X700 = 217 



^ = ^ = 21. 7, for which ^= 1. 3. 



At a distance of 13 feet from the top we can increase the spacing to 



12 inches. 



1" 

 The area of ^ rounds 12" on center = 0.20 square inch. 



Z = 0.20X700 = 140. 



y^ = -yry = 14, for which -j^ =0.6. 



