DESIGN OF FIXED ENDED ARCHES BY THE ELASTIC THEORY 145 



interscepts of the triangle nn'K. is an equal distance from the left 

 springing point. The lines UW and XY contain these centroids 

 respectively. 



Next find the horizontal distance x' ' ' between the lines UW and 

 XY and also their distances x'" and x" from ET. Thus .-//"= 14.957 

 feet, «^=16.063 feet, and x' ' ' =31.02 feet. 



Bm is the left end ordinate of the true closing line tain' of the 

 trial polygon wtBIJK?/!', Km' is the right end ordinate. This 

 line mm' makes 2Ma?=0 in the trial polygon in addition to SM=0 

 as already accomplished by the line nn ' . 



x'" 



Km = __ 1- — 



x'" 



The distances Bm and Km' and therefore the position of the 

 closing line mm' can be found also by graphics. On the line RT lay 

 off E,T=B;?=K7i' =23.524 feet. Assume a pole Q and draw rays 

 to the points K and T. QT cuts XY in a point Y. Through T draw 

 TW parallel to QR to cut the line WU in W. Connect W and Y 

 and through the pole Q draw QS parallel to WY to cut the line 

 RT at S. Scale the distances RS and ST. Then locate the closing 

 line mm' by the relations: — 



211.71 



:RS::B7i:Bm. 



18 

 211.71 



lb 



:ST.:B/?,:Km' 



211.71 



The fraction ^ — is the average of the figures in column 6. 



18 5 5 



The trapezoid mm' KB is now determined. Its interscepts are next 

 desired. Draw the diagonal inK. dividing the trapezoid into the tri- 

 angles mKB and mm'K. Find the interscepts in these triangles by 

 methods similar to those by which the interscepts of the triangles 

 /I.KB and 7?7?/K were found. The results are given in columns 8 



