266 



STRESSES IN A BALTIMORE TRUSS. 



Art. 136. 



D' 8 = (f| X40 ; 000 - i X 33,000) sec =68,500 Ibs. ( -) 



as counter, max. 



D' 7 = ( fJX 40,000- 1 X 33,000) sec 0= 25,000 Ibs. (-) 

 D' = Q-fx 40,000 -21 X33,000) sec = no reversal as counter. 

 V s = JX 33,000 +40,000 =62,000 Ibs. (-) 

 V' s = iX33,000 = 11,000 Ibs. (+) 

 7 X = 1^X33,000 + 1JX40,000= 98,500 Ibs. 

 V 2 =4J X33,000 +1^X40,000 =310,500 Ibs. 

 7 3 =2iX33,000+ff X40,000 = 189,500 Ibs. (+) 

 F 4 = JX33,000+fX40,000 = 81,000 Ibs. (+) 



with counter. 

 F 4 = 11X33,000 + 1 X40,000= 84,000 Ibs. (+) 



symmetrical load, max. 



t/ 3 = U.i = 12i (33,000 +40,000) tan = 730,000 Ibs. ( +) 

 U 5 = U = 151(33,000 +40,000) tan =905,200 Ibs. ( +) 

 U 7 = U 8 = 16-1(33,000+40,000) tan =963,600 Ibs. ( +) 

 LI =L 2 = 71(33,000 + 40,000) tan =438,000 Ibs. (-) 

 L 3 =L 4 = 7 (33,000+40,000) tan 0=408,800 Ibs. (-) 

 L 5 =L G =12 (33,000+40,000) tan 0=700,800 Ibs. (-) 

 L 7 =L S = 15 (33,000 +40,000) tan =876,000 Ibs. (-) 



Fig. 193 shows a Baltimore truss of 14 panels with sub- 

 diagonals all compression members; the other diagonals (except- 



Fig. 193. 



ing end posts) are tension members. The sub-panel loads are 

 carried to the main truss as shown in Fig. 190 (c). 



The coefficients for dead load will be gotten first. For the 

 diagonals in the panels having but one diagonal, the coefficient 

 is equal to the shear in the panel; for the other diagonals, the 

 algebraic sum of their vertical components must equal the shear 

 in the panel. Panel 12-13 is similar to the second panel in 

 Fig. 191. The shear in panel 11-12 is 61-2=41; since Z)' 3 

 carries a half panel load of shear, D 3 must carry 4, because both 

 vertical components act downward. The shear in panel 7-8 is 

 1 and this ia resisted by D' lt making the coefficient D 7 zero. 



