QUANTITIES OF MATERIALS 281 



(b) The thick outer part of the base of the silo (i.e. the footing for the 



wall) may similarly be treated as a hollow cylinder of external 

 diameter, 18' 4"; internal diameter, 11' 0"; height, V 0". 



(c) The thin inner part of the base (i.e. the floor of the silo) may be 



looked upon as a solid cylinder of diameter, 11' 0"; and 

 height 4 inches. 

 (a) 15' 10" = 15|-[}' = 15f = -*£ feet. 



Let us call the volume of the external cylinder Vc, then — 



ttX 95X95 

 Xe ~ 4X6X6 X °° 



by substituting in formula (I.), hence — 



Ye = 5904 cubic feet. 



Again substituting in formula (I.), and calling the volume of 

 the internal cylinder Vi, we have — 



.X 15X15 



4 



Vi = 5299 cubic feet ; 

 hence— 



Volume of wall = Ye — Yi 



= 5904 - 5299 

 — 605 cubic feet. 



(b) 18' 4" == lS r V = 1S.V = 4f feet, hence, in this case— 



, f 7T X 55 X 55 



Ve = ^xinnr xl 



Ye = 264 cubic feet. 



and- Yi = ° Xl ' X "xl 



4 



Yi = 95 cubic feet ; 



hence — 



Volume of thick outer part ) 



\ = Yc — Vi 

 of base (i.e. wall footing) J 



= 264 - 95 



= 169 cubic feet. 



(c) 4" = T y = l foot 



therefore— V~= * X X \ X U X 1 



4 3 



V = 32 cubic feet nearly ; 



