BENT WITH SIDE SHEDS 105 



and O-C, and is equal to C-B (line C-B is not drawn in force polygon) 

 in amount. The reactions R and R f are calculated by the graphic 

 method as previously described. 



The calculation of stresses is begun at point B in the windward 

 column, and in the stress diagram the stresses at B are found by 

 drawing the force polygon a^B-A-b-a. The remaining stresses are 

 calculated as for a simple truss. In calculating the stresses in the 

 ventilator it was assumed that diagonals 9-10 and 10-12 are tension 

 members, so that 9-10 will not be in action when the wind is acting as 

 shown. Before solving the stresses at the joint 6-7-9 ^ was necessary 

 to calculate the stresses in members i-n, 10-11, and g-h. The re- 

 mainder of the solution offers no difficulty to one familiar with the 

 principles of graphic statics. 



The stress in post b-a is equal to V , while the stress in i-c is found 

 by extending i-c to c' in the stress diagram, c' being a point on the load 

 line. The stress in post n^A is equal to V , while the stress in 19-^ is 

 found by extending ig-m to m r in the stress diagram, m' being a point on 

 the horizontal line drawn through C. The kind of stress in the different 

 members is shown by the weight of lines in the bent diagram and by 

 arrows in the stress diagram, one arrow indicating the direction and kind 

 of stress the first time a stress is used and two arrows indicating the 

 second time a stress is used. 



TRANSVERSE BENT WITH SIDE SHEDS. Transverse 

 bents with side sheds are quite often used in the design of shops and 

 mills. The calculation of the stresses due to wind load in a bent of this 

 type is an interesting application of the author's graphic solution of 

 stresses in transverse bents. 



It is required to calculate the stresses due to a horizontal wind 

 load of 30 Ibs. per square foot on the sides and the normal component 

 of 30 Ibs. (Hutton's Formula, Fig. 6) on the roof, the bents being 

 spaced 20' o" centers, as in Fig. 57c. The loads are calculated, and by 

 means of a force polygon in (d) and an equilibrium polygon in (a) 

 the resultant wind 3 W is found to pass through point E, and to be 

 equal to 30,800 Ibs. 



Calculation of Reactions. The horizontal shear of 25,400 Ibs. 

 will be taken by the columns in proportion to their rigidities, in this 



