Art. 131. STRESSES IN A SIMPLE WARREN TRUSS. 



253 



decreased by one-sixth, as may readily be shown by taking suc- 

 cessive sections. 

 Chord Stresses. 



L,=2P tan 0=2X2,250= 4,500 D. L. 

 = do =2X7,875= 15,750 L. L. 



-20,250~ Total. 



J7 1= 4P tan 0=4X2,250=+ 9,000 P. L. 

 = do =4X7,875=+31,500 L. L. 



+40,500 Total. 



L 2 =5 P tan 0=5X2,250= 11,250 D. L. 

 *= do =5X7,875= 39,375 L. L. 



50,625 Total. 



_ L 3 =+Z7 2 =6P tan 0=6X2,250=13,500 D. L. 

 = do =6X7,875=47,250 L. L. 



60,750 Total. 

 As a check, taking the total panel load 13,500, 



Jlf 8 =0 or 2JX15XJ2J- 13,500 (li+i) 15 = 10 L 3 . 

 X 15 X 27,000- 27,000 X 15) =60,750 Ibs. 



It should be noted that the horizontal components of the 

 diagonal stresses all act toward the right at the upper joints and 

 toward the left at the lower joints; hence the chord coefficients 

 are gotten by the addition, at each joint, of all the diagonal 

 coefficients. 



132. Stresses in a Subdivided Warren Truss. Fig. 

 186 shows a subdivided Warren truss of eight panels. This 



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- 8 Panels at 20 ft. =160 ft. 



Fig. 186. 



is a very common type for riveted bridges up to 200 ft. span. 

 The coefficients are given on the truss diagram and are obtained 



