362 



PRACTICAL MATHEMATICS 



If p is the intensity of the breaking load per square inch of 

 section, 



W 



Then 

 and 



P = 



f 



f 



When the ends are free c= 1 and C = -4= 



7i Hi 



f 

 J 



When the ends are fixed c = 4 and C = 



When one end is fixed and the other end is free c = 2-047 and 



2-0477; 2 E 



Example. A hollow cast-iron column 24 feet long has to carry 

 safely a load of 50 tons. The external radius is 5 inches, find the 

 internal radius, (1) when the ends are free, (2) when the ends 

 are fixed, (3) when one end is free and the other fixed. /=- 36 

 tons per sq. in., E = 6000 tons per sq. in., factor of safety = 6. 



Let x be the internal radius. 



Area = 7c(25 a? 2 ) sq. in. 



Moment of inertia = (625 # 4 ) inch units 

 4 



_ 25 + a 2 



4 



Allowing for the factor of safety, the column must be designec 

 to carry a load of 300 tons. 



Then 



300 



7T(25- 



(1) When the ends are free, 

 36 



tons per sq. in 



C = 



and 



7C 2 x 6000 



300 



7t(25 - x 2 ) 



6-08 x 10-* 



36 



1 + 



6-08 x 10- 4 x (288) 2 x 4 



25 



36(25 + a? 2 ) 

 226-8 + x 2 



