CHAP. XH.] METHODS COMBINED. 169 



The construction, then, is simply as follows : Take any point 

 on the direction of P, and draw P D, P E through the points 

 of inflection. Lay off P to the scale of force as A P, and draw 

 A O parallel to P E or P D. We have thus the pole distance 

 H, and the shearing forces P H and H A at B and C. 



B D or C E to the scale of distance, multiplied by H to the 

 scale of force, give the moments at B and C. That is, D and 

 E may be regarded as the points of application for H. The 

 forces along P D and P E considered as acting at these points 

 are held in equilibrium by the reactions P H and H A = L 

 and 1 L and H. Since H, acting as indicated in the figure 

 with the lever arm B D or C E, causes tension in the upper 

 fibres, the moments at B and C are positive. 



2d. Unloaded Span Fig. 78, PI. 21. 



As we have already seen in Art. 93, the inflection points in 

 the unloaded' spans are independent of the load, and are found 

 by the simple construction there given for the "fixed points? 

 Since each fixed point lies within the outer third of the span, 

 we have in Fig. 78 the broken line a b e, referred to in Art. 94, 

 where the moments are alternately positive and negative, and 

 increase from the end, so that any one is more than twice the 

 preceding. Lines drawn parallel to these lines in the force 

 polygon, cut off from the force line the reactions at the sup- 

 ports. Thus, in Fig. 78, c b in the force polygon gives the 

 reaction at D, a b the reaction at C, and if B were an end sup- 

 port that is, if b a went through B a H would be the re 

 action at B. For the resultant shear at D, we should then have 

 a H ab+-cb = Hc. So for any number of spans ; the in- 

 flection points in the loaded span being known, we can easily 

 find infixed or inflection points in the other spans, which are 

 independent of the load, and depend only on the length of 

 these spans. Then draw the broken line a b c P d. Then find 

 the pole distance H by laying off c P = P to scale, and draw- 

 ing cO parallel to cP, and through the point O thus deter- 

 mined drawing H O. Then find the reactions at the other 

 supports, or the shear at any support, by lines in the force 

 polygon parallel to a b, b c, etc. Thus the shear at B k the 

 distance H a cut off by H and O a parallel to a b. Since the 

 shear at D is plus and alternates from D, we have at B the 

 shear + H a. The shear at C is H b ; at D, H- H c, etc. O c 

 being parallel to PC; O a, to b a ; O b, to c b, etc. 



