352 NOTE TO CHAP. I. [APPENDIX 



any number of forces are in equilibrium, the force polygon is 

 closed. Inversely, then, a closed force polygon indicates forces 

 which, if applied at a common point, would hold each other in 

 equilibrium. 



Thus Y 3, 34, X 4, and the weight are, or would be, if all 

 applied at a common point, in equilibrium. This we see 

 directly from the Fig. Tims we know that when any num- 

 ber of forces are in equilibrium, the algebraic sums of their 

 vertical and horizontal components must be zero, otherwise 

 there must, of course, be motion. Now the vertical component 

 of Y 3 plus that of 3 4 minus that of X 4 is exactly equal and 

 opposed to the weight, while the horizontal component of Y 3 

 plus that of 3 4 is equal and opposed to that of X4, and there 

 is then equilibrium. 



Again, according to the principle of Art. 5, Chap. I., any 

 line, as the one joining 2 and 6 (broken line in Fig.), is the 

 resultant of X 2 and X 6, as also of 2 3, 3 4, 4 5 and 5 6. 



The Fig. also well illustrates the points to be avoided in 

 making a strain diagram, already alluded to in Art. 13, Chap. 

 I. The scale to which the frame is taken is here altogether 

 out of proportion to the scale of force. The first should be 

 increased or the second diminished, or both. The present 

 length of the diagonals and flanges is inadequate to give 

 with sufficient accuracy the directions of strain lines of such 

 length. 



Nevertheless we have experienced no difficulty in checking 

 to tenths of a ton the results given by Stoney for this structure. 



5. In PI. 1, Fig. II., we have represented a roof truss, span 

 30 ft., rise 8 ft., camber 1 ft ; and the strain diagram illustrates 

 in its two symmetrical halves (one full, the other dotted) the 

 remarks of Art. 13, Chap. I., upon the check which in such cases 

 our method furnishes of its accuracy. 



We lay off the weights 1, 2, 3, 4, 5, and then the reactions at 

 A and B, which should bring us back to the point of beginning, 

 and thus complete the force polygon. The strains are then easily 

 found, and the two halves should be perfectly symmetrical, and 

 give the same results. 



In Fig. III. we have given another form of truss with strain 

 diagram, the other half of which the reader can complete and 

 letter for himself. 



<* In Fig. IY. we have a form called the French roof trass 



