STRESS DIAGRAMS 193 
It is obvious that the triangles (m) and (0) may be applied to the 
triangle (m) so as to form the figure (7), and this figure gives all the 
‘results which were found from the separate triangles (7m), (m), and (0), 
_and this figure (r) is the complete stress diagram for the given framed 
‘structure. ‘The figure (r) may of course be drawn at once without 
drawing the triangles (m), (n), and (0). It should, however, be noticed that 
in order that the force polygons for the different joints may be combined 
into one diagram, these polygons must be drawn by taking the forces in the - 
order in which they occur in going round each joint in the same directiun. 
_ (r) is the form of the stress diagram when the forces are taken in the 
order in which they occur when going round each joint in the watch- 
hand direction, and (w) is the form of the diagram when the order is 
_ reversed. 
3 178. Example.—A roof truss carrying a load at each joint is shown 
in Fig. 263. The loads are in pounds. The total load is 10,000 lbs., 
and since the loads are placed symmetrically about the centre of the truss, 
- it is obvious that the reaction at each support is 5000 lbs. In cases 
_ where the loading is not symmetrical, the reactions at the supports may 
_ be determined by means of a funicular polygon. 
, 
drawn. The polygon of forces Jmnhkl for the joint LMNHKL may now 
be drawn, and for practical purposes no more of the stress diagram need 
N 
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