102 



RESISTANCE OF MATERIALS 



179-, the factor of safety 



moment which would give the same factor of safety. Finally, com- 

 bine this equivalent bending moment with that due to the eccentric 

 load, and calculate the unit stress from the ordi- 

 nary beam formulas. 



To illustrate the method, suppose that a col- 

 umn 18 ft. long is composed of two 12-in. I-beams 

 each weighing 40 lb./ft., and carries a column load 

 of 20 tons at its upper end and also an eccentric 

 load of 10 tons with eccentricity 2 ft., as shown 

 in Fig. 84. Assuming that the column has flat 

 ends, and using Johnson's straight-line formula, 



P = A ^52,500 



against column failure is 



^52,500 - 



% 



FIG. 84 



= 2 (11.76) (52,500 -179 (47.3)) 

 60,000 60,000 



Now consider the column as a beam and find the equivalent central 

 load K corresponding to the factor of safety just found, namely, 

 17.3. The maximum moment in a simple beam bearing a concen- 



Kl 



trated load K at the center is M = 



Hence, from the beam 



formula M- we have = *--; whence K { Assuming 

 e 4 e le 



the ultimate strength of the material to be 60,000 lb./in. 2 , we have 

 60,000 



lb./in. 2 , 7-2(245.9) in. 4 , 

 e = 6 in., 



17.3 

 I = 216 in., 



and, inserting these values, the equivalent load K is found to be 

 4 x 60,000 x 491.8 



17.3 x 216 x 6 



= 5220 Ib. 



Now the eccentric load P z , acting parallel to the axis of the column, 

 produces the same bending effect as a horizontal reaction H at either 

 end, where HI = P 2 d. The bending moment at the center, due to a 



If I 

 reaction H perpendicular to the axis of the beam, is, however, . 



A 



