Concrete Silos 53 



D 



( )2x3.14xHx30- pounds of silage. 



2 

 Where D equals diameter of silo in feet. 



H equals depth of silage in feet. 



30 equals weight of silage per cubic ft. 



EXAMPLE: To get the capacity of a silo ten feet in diameter 

 containing twenty feet of silage. 



10 

 (-)2 X 3.14 X 20 X 30= 



52x3.14x20x30= 



5 X 5 X 3 - 14 X 2 X 30 = 47 > 100 pounds silage. 

 47,100-=-2,000=23.5 tons of silage. 



In the eighth annual report of the Wisconsin Agri- 

 cultural Experiment Station, Prof. F. H. King gives 

 the results of investigations to determine the pressure 

 of silage against the silo wall. It was found in these 

 experiments that the pressure of silage upon the silo 

 walls increases with the depth and is equal to 11 

 pounds per square foot of each foot of depth. Thus at 

 a depth of 20 feet, the bursting pressure in a silo is 

 220 pounds per square foot, and at a depth of 35 feet 

 the pressure amounts to 385 pounds. 



Prof. J. B. Davidson, of the Iowa Experiment Sta- 

 tion, Ames, la., says that a careful investigation of 

 modern practice proves that an allowance for this pres- 

 sure is sufficient, and that many concrete silos are now 

 standing and in successful service with much less re- 

 inforcement than that required by an assumed pressure 

 of 11 pounds per square foot per foot of depth. This is 

 due to the fact that the wall independent of the steel 

 is able to resist a part of the bursting pressure. 



Table IV, prepared by Prof. Davidson, gives the 

 steel required in cylindrical silos to carry a bursting 

 pressure of 11 pounds per square foot per foot of depth 

 and it is based on a safe tensile strength of 20,000 



