SIMPLE STKUCTUKES 



189 





Since R is by assumption equivalent to the combined action of 

 the two shear legs, the thrust T in each may be found by resolving 

 forces along R. Thus Tcos a = J R, which determines T, since R 

 has already been found. , 



Similarly, the force at 

 the bottom of the shear 

 legs tending to make 

 them spread is T sin a. 



At the point C the 

 forces acting are the up- 

 ward pull V on the an- 

 chorage, the horizontal 

 pull H on it, and the 

 tension P in the guy. 

 Hence, applying the con- 

 ditions of equilibrium, 

 we have 



H=Pco$l3,y=P sin 0. a 



2. Graphical method. 

 To illustrate the graph- 

 ical calculation of stresses 

 from joint reactions, con- 

 sider the roof truss 

 shown in Fig. 127. 



Since the loading in 

 this case is symmetrical, the reactions of the supports will each 

 be equal to half the weight on the truss. 



The most convenient notation is to letter the spaces between the 

 various lines of the diagram. Each member of the truss and each 

 external force will then be designated by the adjoining letters on 

 opposite sides of it, as the member AH, the load BC, etc. 



Starting with the left support, we have three forces meeting at 

 a point. The magnitude of one, namely R^ or AB, is known, and 

 the directions of all three are known. Hence the other two can be 

 determined by means of a triangle of forces. Thus, if ab is laid off to 

 scale to represent R^ and aj\ bj, are drawn from a and b parallel 



FIG. 127 



