COLUMNS AND STRUTS 103 



Hence the total equivalent moment at the center now becomes 



Kl HI _ Kl P z d 



: 4 "" 2 == 4 " 2 



5220 x 216 20,000 x 24 





= 521,880 in.-lb. 

 'Consequently, the maximum unit stress in the member becomes 



which corresponds to a factor of safety of about 9. 



If this factor of safety is larger than desired, assume a smaller 

 I-beam and repeat the calculations. 



A method substantially equivalent to the above is to assume that 

 the stress in a column is represented by the empirical factor in the 

 column formula used. Thus, for a short block the actual compressive 

 stress p is given by the relation P = pA, whereas in the column 



formula used above, namely, P = A( 52,500 179 - ) , the stress p is 



I t ' 

 replaced by the empirical factor 52,500 179 - Consequently, the 



fraction 7 



52,500-179- 



where u c denotes the ultimate compressive strength of the material, 

 represents the reduction in strength of the member due to its slim- 

 ness and method of loading ; or, what amounts to the same thing, 

 the equivalent unit stress in the column is 



(136) 



^52,500-179- 



\ 



Applying this method to the numerical problem given above, we 

 have A = 23.52, 



- = 47.3, 

 t 



and 



60,000 



52,500-1791 52 ' 6 ( 



t 



= 1.36. 



