588 



THE DESIGN OF STEEL DETAILS. 



CHAP. XVII. 



To locate the neutral axis A-A , take moments about the lower edge of the channels 



8.92 X 5 +4-38 X 10.156 



13-30 



= 6.70 in. 



The eccentricity is e = 6.70 5.00 = 1.70 in. The moment of inertia I A> about axis A-A 

 may be calculated as follows: 



Let I c = I of channels about center of channels (Table 14). 

 IP = I of plate about center of plate (Table 4). 

 AC = area of channels (Table 14). 

 A p = area of plate (Table i). 



V--7400 



a =3.87 

 = 12.50' 



FIG. 12. END-POST OF A HIGHWAY BRIDGE. 



Then I A = I e + I p + A c X i-7o 2 + A P X 3456 2 . 



= 2 X 66.9 + 0.04 + 8.92 X I-70 2 + 4.38 X 3456 2 

 = 133-8 + 0.04 + 25.76 + 52.20 

 = 211.80 in. 4 



Then r A = ^I A + A = ^2 11.80 -r- 13.3 = 3.99 in. 



The moment of inertia IB, about axis B-B may be calculated as follows. 

 Let I c ' = I of channels about neutral axis parallel to the web (Table 14). 

 IP' = I of plate about vertical axis (Table 3). 

 A c = area of channels (Table 14). 



From Table 82 the distance back to back of channels is 8% in. From Table 14 the distance 

 from neutral axis to back of channel is 0.639 in. The ^'stance from neutral axis of channels to 

 axis B-B is 4.25 + 0.639 = 4-889 in. (4.89 in. will be used). 

 Then I B = // + /' + A c X 4-9 2 



= 4.60 + 71.46 + 9.82 X 4.8g 2 

 = 4.60 + 71.46 + 213.28 

 = 289.34 in. 4 __ _ 

 Then r B = ^I B *- A -. "^289.34 -*- 13.3 = 4.67 in. 



