CHAP. Vin.] CONTINUOUS GIEDEKS. ] 39 



2d. If we take \ = I, we have at once M' = y' and M" = y" 

 In the inner spans, therefore, we have directly, as we have al- 

 ready seen, for the special case (Art. 72), when \= I, the mo- 

 ments at the supports. 



3d. For a span adjoining a span whose length is \, we have 

 the moment for the intervening support directly from this last 

 span. If the inner spans have the same length I, and the two 

 outer the same length l, we can accordingly, by making \ = I, 

 obtain directly the moments at the supports. 



9O. Comparison with Girder fixed horizontally at both 

 ends. If the ends are fixed horizontally, the lines in the force 

 polygon parallel to A U and B V coincide, i.e., H and G fall 

 together. Accordingly, if we designate the end moments now 

 by m f 2ft" we have (Fig. 53) 



|(FG + GE) = lpEor^ (2ft'+3ft") = 2ft. * 



Therefore, in the first equilibrium polygon, the moment areas 

 on each side of the closing line are equal. 



Indicating the points for this case by the index [Fig. 54, 

 PI. 15], we have the triangle A U M equal to U V V , 

 and therefore A M = V V' = V V, as also, in like manner, 



B N = U U'o = U U', or, taking b = 1 \, 



U U' = 2ft' l-\ 2 V V = 2ft 



Therefore, the ordinates between the cross-lines at the verti- 

 cals passing through U and V are, for gvrders fixed horizon. 

 tally, proportional to the end moments 2ft' and 202". 



For \ = l t b = - I, and these ordinates give the moments 



directly. 



If we draw through U' and V a straight line intersecting 

 the end verticals in Q and R, and prolong N U' and M V to 

 intersections S and T, then Q M = 2 U U', Q S = V V, and 



hence M S = 2 U U'+V V = (2 m'+W) (-V; and in the 



same way N T = (2 W+W) 



Therefore, the segments cut off upon the end verticals by the 

 cross-lines are proportional to 2 3)^+ 2ft" and 2 2ft" + 2ft'. 



