188 GRAPHIC AND ANALYTIC [CHAP. XH. 



2374 ?-0.1836 



D = -P fo.2374 ? 

 \ I 



KT a . 1 2 3 4 , 



iNow - is -, -, -, jiths, etc., according to the position of the 

 l> 7777 



& 8 1 8 27 

 weight at 1st, 2d, 3d apex from end. So also ^ is , , - 



fr O'lO &XO 



etc. The above equations for the reactions, then, may be 

 written A = P (1-0.1614 b + 0.000535 b s ), 



B = C = P (0.09766 b- 0.000535 J 8 ), 

 D = -P (0.03391 5-0.000535 I s ), 



where b has the values 1, 2, 3, 4:, etc., for P x , P 2 , P s , P 4 . 



Thus, if we wish the reactions due to a weight P 4 of 9 tons at 

 the fourth apex, as shown in Fig., we have only to make P = 9 

 and b = 4, and we find at once A = 3.498 tons, B C = 

 3.207 tons, D = 0.912 tons. The sum of all these reactions 

 exactly equals P, as should be. 



The middle supports are supposed raised by an amount equal 

 to the end deflections of the open draw, therefore the strains 

 due to dead load are easily found, as in the " braced semi-arch," 

 Art. 9. 



The reactions due to live load, according to Art. 121, will not 

 be affected by this raising of the supports. 



To find the strains due to P 4 , we draw the force line E 2 P 

 [Fig. 86 (a)~\ by laying off P 4 = 9 tons down from F to F 1? then 

 F! E 2 downwards equal to the negative reaction at C, viz., 

 0.912 tons. Then from E 2 lay off upwards Eg E x = to positive 

 reaction at C =+3.207 tons. Then E : E = reaction at B = 

 + 3.207 tons, and finally E F equal to reaction at A = +3.498 

 tons, which should bring us back exactly to point of beginning 

 F, since the reactions and the weight P must be in equilibrium. 

 [Note. "When we wish to begin at the left end of the frame, 

 it is best, as in this case, to lay off the reactions in order, com- 

 mencing at the right.] We have taken the scale of force 4 

 tons per inch. 



The weight P 4 acts upon the triangulation drawn full in the 

 figure. Using now the notation of Art. 114, and representing 

 all the space above the truss by E, all below by F, we have at 

 A the reaction E F [Fig. 86 (a)] in equilibrium with E 1 and 

 F 1, and drawing parallels to these lines from E and F, we find 

 the strain in F 1 = 3.54 tons tension, and E 1 = 5.1 compression. 



