148 CONTINUOUS GIRDER. [('II A P. X. 



end tangents and the closing line A' B', the distances on the 

 force line are the reactions at the ends of the span. Instead 

 of this we may lay off from A and B, A G and B H equal to 

 the first pole distance a, and through G and H draw parallels 

 to the end tangents, intersecting the verticals through A and B 

 in A' and B^ We thus obtain the reactions A A t , B B^ The 

 ordinates to A l B L then give the shearing force at any point. 



If the line p\ 2 representing the moment units is equal to m, 

 and that representing the force units p \ = n, then the first 



pole distance must be a = *-r* A. = \. 



* m 



Accordingly, it is now easy, from the general construction 

 given in Art. 97, to construct the shearing force and moments 

 for uniform or dead load of girder in any case. Let us pass 

 on to an example illustrating more fully the above principles. 



99. Example. As an example of the application of the 

 above principles, we take a girder of four spans, as given in 

 PI. 17, Fig. 63. The two interior spans are each 96 ft., the 

 exterior spans 80 ft. each ; that is, ^ : I : : 5 : 6. Choose any 

 scale of length convenient, as, for instance, 50 ft. to an inch, 

 lay off the spans and construct first the fixed points. For this 

 purpose we draw the third and limited third verticals. These 

 last are easily found from the principle already deduced, that 

 they must divide the distance between the third verticals into 

 segments inversely as the corresponding spans [Art. 92]. Lay- 

 ing off, then, from the third vertical in the first span, I to the 

 right, or from the third vertical in the second span 4 to the 

 left, we have the first limited third vertical. The same at the 

 other end gives the other. For the centre support, of course, 

 the limited third, since the adjacent spans are equal, passes 

 through the support itself. We can, therefore, now construct 

 the fixed points according to Art. 93. 



Let the load per unit of length p be \ ton per ft. Then 

 taking \ [Art. 88] equal to I, we have n = p\ pi = 48 tons 

 and m = p\ z =p#= 4608 ft. tons [Art. 98 (3)]. It remains 

 to assume a scale of force. Let this be 20 tons per inch, then 

 our moment scale is 20 x 50 = 1000 ft. tons per inch. The 

 values of which we shall need to make use are, then, to scale 

 li = 1.6 inches, X == I 1.92 inches, 



= 0.6944 inches, = 1.44 inches, = 0.4823 inches 



