CHAP. IX.] LOADED AND UNLOADED SPANS. 143 



The construction of the fixed points is the first operation in 

 the graphical treatment of the continuous girder. 

 The above construction was first given by Mohr. 



94. Shearing Force, Reactions at tbe Supports, and Mo- 

 ments in the Unloaded Spans. The moments in the unloaded 

 spans are given, then, by the ordinates to a broken line whose 

 angles lie in the support verticals, and which, for the case of 

 supports on a level, passes through the corresponding fixed 

 points. 



4t follows directly that the moments at the supports are 

 alternately positive and negative, and increase from the 

 end, so that any one is more than three times the preceding. 

 (See Art. 111.) 



Since now this polygon has alternately angles down and up, 

 the reactions at the supports must be alternately positive and 

 negative. From the corresponding force polygon it follows 

 that they must increase from the end. 



The shearing forces are,, therefore, also alternately positive 

 and negative, and increase from the end on. 



95. Loaded Span. Let now the span A B [PI. 15, Fig. 58] 

 be arbitrarily loaded. It can be proved here also, as in Art. 

 92, that the prolongation of the sides U' V and S U, as also of 

 V" U" and S V, intersect in the limited third verticals. 



When the supports are in a straight line, then, by the con- 

 struction of Art. 92, the fixed points I and K are the intersec- 

 tions of S V and S U with A B. We can, therefore, at once 

 assert, that the sides S U and S V of the second equilibrium 

 polygon pass through the fixed points I and K, when the sup- 

 ports are on a level. 



For known position of the fixed points and for given load, it 

 is, therefore, easy to draw the second polygon by drawing ver-. 

 ticals I I x and K K x equal to the corresponding ordinates of the 

 cross-lines. S U and S V pass, then, through I Kj and K I t re- 

 spectively. Then, by Art. 89, the moments at the supports may 

 be determined. 



Since A I < - 1, U must lie to the right of I, and the angle 

 o 



A US is concave downwards. Accordingly, the force at U, 

 viz., - M' I, acts upwards. The same holds good for V. Hence, 



