EI.ISHA MiTCHELIv SCIEJNTIFIC SOCIETY. 48 



abutment (14 — p') — (6.333^S) = 7.666, just as we should 

 find for the two orig-inal loads b}^ the law^ of the lever. 



These results are true for any value of ^(< 14) that 

 may be assumed. The^^ do not determine p. We shall 

 soon see how to determine it after expressing- the work 

 of deformation of the entire truss iu terms of -p. 



As the elastic work in the vertical posts is very small 

 it will be neg-lected. The work in the ties s given by 

 formula (Ij. 



For the two inclined ties, the stress, 5 = 15.5 x (14 — ■/) 

 -^4, a=\S.S and zt'=.04 square foot nearly. Substituting 

 these values in (1) and multiplying- by 2 we get the elas- 

 tic work by inclined ties at either end. 



Similarly the stress in the horozontal tie CC, ^=15 

 (14 — ^)-4-4, also a — Vb and w =.04. These values substi- 

 tuted in (1) g-ives the work of horizontal rod CC. 



As the modulus e is for iron and E for wood, take 

 ^=16 E, then the work of the iron ties in one truss will 

 add up to, 



1 



(1075/^—29600/) (1). 



2E 



on neg-lecting" terms that do contain /, as will be done in 

 what follows, as snch constant terms disappear when dif- 

 fertiated as to p. 



To ascertain the work on the chord acting- as a beam, 

 we use formula (2). We have the down waad force / act- 

 ing- at left post, the upward force (9.5 — f>) at right post, 

 the upward reaction at left abutment=(/ — 3.166) and 

 the downward reaction at rig-ht abutment ^(6. 333 — -/). 

 The beam can thus be considered as free and subjected to 

 these forces only, 



Let us apply formula (2) first to right segment of 

 chord from E to B, 15 feet in leng-th, 

 .-. 2/= 15, 



Mo = 0, 



Mi-= (6,333— /)7.5, 



