ic6 STRESSES IN A TRANSVERSE BENT 



case the rigidities of the columns are assumed equal and the shear at 

 the foot of each column will be 6350 Ibs. The vertical reactions will 

 be due to two forces: (i) a vertical load of 17,200 Ibs., which will be 

 taken equally by the four columns, making a load of 4300 Ibs. on each ; 

 and (2) to a bending moment of 25,400 Ibs. X 9- 2 ft - = 2 33>68o ft.-lbs. 

 (the bending moment about C. G. is also equal to 30,800 Ibs X 7-6 ft.), 

 which will be resisted by the columns and will cause reactions varying 

 as the distance from the center of gravity of the columns, (E), as in the 

 case of the continuous portal, Fig. 63. 



Let v\, v' z , v' z , and z/ 4 represent the reactions due to moment in 

 the columns, respectively; then if a is the reaction on a column at a 

 units distance from the center of gravity we will have v\ = a 40, 

 7/ 2 = a 20, z/ 3 = + 2 > an d v\ = + a 40. The resisting moment 

 of each column will be equal to the reaction multiplied by the distance 

 from the center of gravity, and a 40* -f a 2O 2 + a 20* + a 40* = 233,- 

 680 ft.-lbs. from which a = 58.42 Ibs. and v\ == 2340 Ibs. ; v\ = 

 1 170 Ibs. ; 7/3 = + 1 170 Ibs. ; v\ + 2 34 Ibs. 



Now adding the reactions due to (i) and (2) we have 



V^ = 4300 2340 = + 1960 Ibs., 

 V 2 = 4300 1 170 = + 3130 Ibs., 

 F 3 = 4300 1 170 = + 5470 Ibs., 

 F 4 = 4300 2340 = + 6640 Ibs. 



Combining the horizontal and vertical reactions we have R^ = a^A = 

 6600 Ibs.; R 2 = A-B = 7200 Ibs.; 7? 3 = J3-C = 8400 Ibs.; R = 

 C-D = 9100 Ibs. These reactions close the force polygon in (d). 



Calculation of Stresses. Auxiliary members are substituted as 

 shown by the broken lines. It will be seen that these members are 

 arranged so that all bending is removed from the columns and that 

 the stresses in the truss members are correctly given in the stress dia- 

 gram. The calculation is started at point A at the foot of the left- 

 hand column as in the case of the simple transverse bent, and reac- 

 tions R 2 and R s are substituted as the calculation progresses, the stress 

 diagram finally closing at the base of the leeward column, point D. 

 The stresses are given on the members in (a). The direct stresses in 

 the columns are easily found by algebraic resolution beginning at the 



