CHAP. Xn.] METHODS COMBINED. 1S1 



is 250 tons compression. It is only necessary, therefc re, to find 

 the strains due to each weight of the live load. From the 

 tabulation we can, then, by means of the ratio of the dead to 

 live load, find the strains due to dead load alone, and then by 

 a comparison of the two find the maximum compressive and 

 tensile strains. If the maximum strains due to live load are of 

 opposite kind, but less than the constant strains due to dead 

 load, we shall need no counterbracing. The resultant strains 

 will then always be of the character given by the dead load. 

 If greater, we must counterbrace accordingly. The process is 

 the same as by the methods of calculation, and the reader may 

 refer to Stoney Theory of Strains for illustrations. 



12O. The " Tipper," or Pivot Draw, with secondary cen- 

 tral Span. We have said that a pivot draw may be considered 

 as a beam continuous over three supports. In practical con- 

 struction this statement needs some modifications which deserve 

 special notice. Thus practically that portion of the beam over 

 the central support forms a short secondary span D D [Fig. 83, 

 PI. 22] the reactions at the supports D and D being always 

 equal and of the same character. If a weight acts, say, on the 

 first span A B, and the beam itself is considered without weight, 

 the end C must be held down, that is, the reaction there is neg- 

 ative. Now as the weight P deflects the span A B (Fig. 83), it 

 causes one secondary support D to sink, and the other to rise 

 an equal amount. In practice D and D may be the extremities 

 of the turn-table, and the reactions are then evidently different 

 from those given by the formulae of Art. 118. 



If in this case we take a as the distance of the weight P from 

 the left support A, the reaction for load in A B will be given 

 by the following formulas : 



Where the ratio = k, a being the distance of the weight P 



if 



from the left support A (for load in the span A D), I = span, 

 A D = D C, and n I = span D D, and where the constant 

 (4 + 8 n + 3 ri*) is put for convenience = H, then 



R A = I 2 H - (10 + 15 n + 3 n 9 ) k + (2 + n) t* I* 

 2 H|_ J 



1^ = 1^:= JU (6 + 9^1 + 3*)*-(2 + n)#J 



* See Supplement to Chap. XIII., Art. 6. 



