558 STRUCTURAL MECHANICS. CHAP. XVI. 



PROBLEM *-, DEAD LOAD STRESSES IN A CAMEL-BACK TRUSS BY GRAPHIC RESOLUTION. 



(a} Problem. Given a Camel-back (inclined Pratt) truss, span 160' o", panel length 20' o", 

 deotri at the hip 25' o", depth at the center 32' o", dead load 400 Ib. per lineal foot per truss. 

 Calculate the dead load stresses by graphic resolution. Scale of truss, i" = 25' o". Scale of 

 loads, i" = 10,000 Ib. 



(b) Methods. The loads beginning with the first load on the left are laid off from the bottom 

 upwards. Calculate the stresses by graphic resolution, beginning at RI and checking up at R^. 

 Follow the order given in the stress diagram. 



(c) Results. The top chord is in compression and the bottom chord is in tension. All 

 inclined web members are in tension; while part of the posts are in compression and part are in 

 tension. Member 1-2 is simply a hanger and is always in tension. 



PROBLEM 2. DEAD LOAD STRESSES IN A PETIT TRUSS BY GRAPHIC RESOLUTION. 



(a) Problem. Given a Petit truss, span 350' o", panel length 25' o," depth at hip 50' o", 

 depth at center 58' o", dead load 0.9 tons per lineal foot per truss. Calculate the dead load 

 stresses by graphic resolution. Scale of truss, i" = 50' o". Scale of loads, i" = 45 tons. 



(b) Methods. The loads beginning with the first load on the left are laid off from the top 

 downwards. Calculate RI and R.2. Calculate the stresses in the members at the left reaction 

 by constructing force triangle i-YX. Then calculate the stress in 1-2 by constructing polygon 

 F-I-2-F. Draw 3-2, which is the stress in member 3-2. Then pass to joint Wi where there 

 appears to be an ambiguity, stress 4-5 being unknown. To remove the ambiguity proceed as 

 follows: At JF 3 on the left side of the stress diagram assume that Ws is the stress in 5-6 (the 

 member 5-6 is simply a hanger and the stress is as assumed). Calculate the stress in 4-5 by 

 completing the triangle of stresses in the auxiliary members. The stresses are now all known 

 at W% except 3-4 and 5~F, but the stress in 4-5 is between the two unknown stresses. First 

 complete the force polygon 2-3-4-5 '-Y-Y-2. Then by changing the order the true polygon 

 2-3-4-5- Y Y-2 may be drawn. This solution is sometimes called the method of sliding in a 

 member. The apparent ambiguity at joint W^ may be removed in the same manner. The stress 

 diagram is carried through as shown and finally checked up at RZ. It will be seen that there is 

 no apparent ambiguity on the right side of the truss. 



(c) Results. It will be seen that the Petit truss is an inclined Pratt or Camel-back truss 

 with subdivided panels. The auxiliary members are commonly tension members in all except 

 the end primary panels as in the Baltimore truss in Problem 6. It will be seen that the stresses 

 in the first four panels of the lower chord are the same. The loads in this type of Petit truss are 

 carried directly to the abutments. The Petit truss is quite generally used for long span highway 

 and railway bridges. 



PROBLEM 3. MAXIMUM AND MINIMUM STRESSES IN A WARREN TRUSS BY ALGEBRAIC 



RESOLUTION. 



(a) Problem. Given a Warren truss, span 160' o", panel length 20' o", depth 20' o", dead 

 load 800 Ib. per lineal foot per truss, live load 1 ,600 Ib. per lineal foot per truss. Calculate the 

 maximum and minimum stresses in the members due to dead and live loads by algebraic reso- 

 lution. Scale of truss as shown. 



(6) Methods. Dead Load Stresses. Beginning at the left end the left reaction is RI = 3-^ \W. 

 The shear in the first panel is 3%W, in the second panel is 2JJF, in the third panel is f TF, and 

 in the fourth panel is \W. Now resolving at RI the stress in i-F = ^W- tan 0, stress i-X 

 = + 3 J IF- sec 0. Cut members i-F, 1-2 and 2-X and the truss to the right by a plane and 

 equate the horizontal components of the stresses in the members. The unknown stress 2-X 

 will equal the sum of the horizontal components of the stresses in i-F and 1-2 with sign changed, 

 = - (- 35 - 3l)JF-tan 6 = + jW tan 0. The stress in 3-F = -(7 + 2|)TF tan = - 

 9iPF-tan 0. Stress in 4~X = - (- 9? - *f)JP'tan = + i2W-tan 0; stress in 5~F = - 

 ( + 12 + i)TF-tan0 = + i&W-tant] and the stress in 6-X = - (- 13! - 1 1) IF- tan = 

 + !5lF-tan0; etc. The coefficients of the chord stresses when multiplied by IF tan give 

 the stresses, while the coefficients for the webs when multiplied by IF- sec give the web 

 stresses. 



Live Load Stresses. Chord Stresses The maximum chord stresses occur when the joints 

 are all loaded, and the chord coefficients are found as for dead loads. The minimum live load 

 stresses in the chords occur when none of the joints are loaded, and are zero for each member. 



Web Stresses. The maximum web stresses in any panel occur when the longer segment into 

 which the panel divides the truss is loaded, while the shorter segment has no loads on it. The 

 minimum live load web stresses occur when the shorter segment is loaded and the longer segment 

 has no loads on it. The maximum stresses in members i-X and 1-2 occur when the truss is fully 



