CHAP. Xin.] ANALYTICAL FORMULA. 209 



Th + + 1?- ' + i? - 



8 ' 388 + 142 ~~ 530' 388 + 142' r 388 -h 142 ~" 530' 



and so on. 



We can, therefore, independently of the theorem and analyt- 

 ical method by which the above results were deduced, produce 

 the table to any required number of spans.* 



134. Total uniform Load all Spans equal Reactions. 

 The moments being known, the reactions at the supports can 

 be very easily found. 



Thus, the reaction at the first or last support is 



U I M -n UI M 



at any other support 



Thus, in our example in Art. 127, we find 



56 161 _ 137 , _ 143 



Hence, the shear at the fourth support is 



56 , 161 . 137 , 143 . 7 71 



l- 3ul = 



71 



or when u I 80 tons, --j^ u I = 40 tons. 



Multiplying this shear by 1.414 (the secant of the angle with 

 vertical), we find for the strain in diagonal a b (Fig. 88) due to 

 uniform load + 56.5 tons, the same nearly as already found in 

 our tabulation. 



135. Triangle for Reactions. The reactions for a number 

 of spans being found, and tabulated, as above, in the case of the 

 moments, we shall have a triangular table precisely similar to 

 the one above, in which the same rule holds good for odd and 

 even numbers of spans within the double lines drawn in the tri- 

 angle. The rule does not apply to the outer rows left and right. 



* The above relations between the moments can be shown analytically to 

 be a result of the properties of the well-known " Clapeyronian numbers." 

 For the table above, as also the others which follow, we are indebted to the 

 kindness of Mr. Mansfield Merriman, Instructor in Civil Engineering in the 

 Sheffield Soi. School of Yale College. They are given, so far as we are aware, 

 in no treatise upon the subject yet published. 



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