Art. 137. 



STRESSES IN A CAMEL-BACK TRUSS. 



269 



after it has been found that the main diagonal stress reverses. 

 To prove that it is true, consider D 3 for example, Fig. 194. 

 Taking a section through U 3 , D 3 , and L 3 , the algebraic sum 



L.,4 L 4 3 

 7 Panels aHO=140 -\ \ 



Fig. 194. 



of the horizontal components of the stresses cut must equal 

 zero, because all external forces are vertical. 



C/3 sin a 3 =Z) 3 sin 3 +L 3 . 



Substituting the values of U 3 and L 3 in terms of the moments 

 at 4 and 9, for whatever loading may be under consideration, 



D 3 sin 6 3 = 



MI sin a 3 M 9 



M 9 



In a similar manner, if D 3 is omitted and the counter tie, D' 3 

 inserted in this panel, 



This proves that if a tensile diagonal is used in a panel of a 

 truss with inclined chords and vertical posts, the horizontal 

 component of its stress is the same as the horizontal component 

 of the stress in the compressive diagonal would be if it were 

 used. 



The stresses will now be calculated for the camel-back Pratt 

 truss of Fig. 194, for the following data: 



Dead Load, 



Trusses and bracing = 600 Ibs. per ft. of bridge. 

 Floor system = 400 Ibs. per ft. of bridge. 



Total 



1000 Ibs. per ft. of bridge. 



