CHAP. I.] COMMON POINT OF APPLICATION. 13 



a and A, and be equal and opposite to the resultant at A. The 

 resultant at the right abutment must pass through that abutment, 

 and also through the intersection of P 2 with A a. So" for any 

 other force, as P 6 , we have simply to draw B a to intersection 

 with P 6 , and then P 6 A. We can now decompose P 6 or P 2 along 

 the resultants through the abutments thus found. Thus resolv- 

 ing P 2 along A a and P 2 B, Fig. 5 (e), we find the force acting at 

 apex a. This force resolved into A and 1 gives the strains on 

 these pieces both compressive. Passing then to the next apex, 

 we obtain the strains in 2 and E. Then to the next, and we 

 get 3 and B, compression and tension respectively, and so on, 

 as shown by diagram, Fig. 5 (V), which, it will be seen at once, 

 is similar to Fig. 5 (/<), already obtained for the " semi-arch," 

 except that the strain in A is less than for the semi-arch and 

 compressive, while B C and D are -in tension. The reason is 

 obvious. At a [Fig. 5 (c)] the resultant lies between A and 1, 

 and therefore causes compression in both, while it passes out- 

 side of the arch entirely, to the right of the apex for diagonals 

 3 and 4, and hence causes tension in B C and D. Fig. 5 (d) 

 gives the strains due to P 6 . Here the resultant or reaction at 

 A is first found and resolved into 9 and H, and then we go 

 through the frame as before. We see that 4: and 5 under the 

 load are both compressed, that E and F are in tension and G 

 and H, as also the entire upper chord, in compression. The 

 work checks from the fact that the line closing the polygon 

 formed by E and 2 should be exactly parallel to and give the 

 strain in diagonal 1, or A and 1 should be in equilibrium with 

 the resultant through a [see Fig. 5 (<#)]. 



In every case of the kind we first, then, have to draw the 

 frame diagram. Then lay off the force polygon which should 

 close. Finally we construct the strain diagram. The frame 

 diagram should be taken to as large a scale as possible consist- 

 ent with reasonable size, and the scale for the force and strain 

 diagrams as small as possible, consistent with scaling off the 

 strains to the requisite degree of accuracy. A small frame 

 diagram does not give with the proper accuracy the relative 

 positions and inclinations of the various pieces, so as to ensure 

 the proper direction for the lines of the strain diagram. A 

 slight deviation from parallelism causes sometimes considerable 

 variation. Nevertheless with practice, care, and proper instru- 

 ments the accuracy of the method is surprising ; even in com- 



