192 GBAPHIO AND ANALYTIC [CHAP. XII, 



therefore the moment at C, is also positive. The lower flange 

 F 9 is therefore compression, or for the dotted system of trian- 

 gulation E 9 is tension. The shear to the left of C, E l F l acts 

 down, since P + A + B = E x F x . Consider F 9 acting at 

 apex 9, and then, since it is compression, it must act towards 9 

 (from right to left), and passing down then from E^ to F l5 we 

 must lay off F 9 to the left of E x F lt For similar reasons, for 

 the other system, E 9 must be laid oif to the right. 



For fixed supports B and C, the moments alternate from B, 

 and the moment at C is therefore negative that is, gives com- 

 pression above and tension below. Flange F 9, for the system 

 of triangulation of P, would then be tension instead of com- 

 pression, as above ; P will, however, still be greater than A + B, 

 and hence the shear is to be laid off down, and F 9 must be 

 laid off to the right. 



If, then, it were required to find the strains in the span B C, 

 preceded and followed, it may be, by many others, it is suffi- 

 cient to know the moment' and shear at one support. "We can 

 then commence and continue the strain diagram, without being 

 obliged to go off to a distant free end and trace all the strains 

 through till we arrive at the span in question. 



126. Method of procedure for any number of Spam. 

 The general method of procedure which we advise, is then as 

 follows. Let us take any number of spans, say seven [PI. 22, 

 Fig. 87], and let it be required to find the maximum strains in 

 the span D E. It is not, as we have just seen, necessary to com- 

 mence at the extreme end A or H, and follow the reaction there 

 through, from span to span, till we arrive at D. As we have 

 seen from the preceding Art, we may start directly from D, 

 provided we know the moment and shear there. Now, since a 

 load in any span causes positive moments and reactions at the 

 two ends of that span, and since either way from these ends the 

 moments and shear at the other supports alternate in character 

 [Art. 102], any and all loads in A B cause positive moments 

 and reactions at D. So also for loads in C D and in F G. 

 Loads in B C, E F and G H, on the other hand, cause negative 

 moments and reactions at I>. [See Fig. 87.] 



To find the maximum positive moment and shear at D due 

 to the other spans, we must then suppose the method of 

 loading shown in Fig. 87 (a). For the maximum negative 

 moment and shear at D, we have the system of loading shown 

 in Fig. 87 (b). 



