CONCRETE DESIGN 289 



If the total load is eccentric, the formulas just given 

 reduce to the following: 

 For circular columns 



For rectangular columns, 



According to these formulas, it is necessary first, in design- 

 ing a column that will stand a given load, to select by 

 inspection the section of column that seems to be about 

 correct, and then to solve the equation for s. This value of 5 

 must be less than the allowable working stress it is pro- 

 posed to use. If it is larger, a larger area of column must 

 be selected, and the problem worked out again; if it is very 

 much smaller, possibly too large a section for economy has 

 been selected, and in this case a smaller section should be 

 assumed and the equation again solved for s. It should be 

 borne in mind that the height of the column must not be 

 greater than twelve times the least dimension of the cross- 

 section. The eccentricity should not be so great as to cause 

 te'nsion in the column. 



EXAMPLE. Design a cylindrical column of 1-2-4 stone 

 concrete, 18 ft. high, to carry with a factor of 6, a central 

 load of 100,000 Ib. and an eccentric load of 100,000 lb., the 

 eccentricity being 4 in. 



SOLUTION. Since the column is 18 ft. high, it should be 

 at least 18 in. in diameter. The ultimate crushing strength 

 may be taken as 2,500 lb. per sq. in. in 6 mo. The safe working 

 stress would therefore be 2,500-*- 6 = 417 lb. per sq. in. 



A column 28 in. in diameter will be tried first. The area 

 of the cross-section is .7854 X28 2 , or 615.75 sq. in. To apply 

 the first formula, A = 615.75, P = 200,000, P e = 100 000, e = 4, 

 and rf = 28. Substituting in the formula, 



200.000 8X4X100,000 



