Loaded Shaft supported in 2 liree Bearings. 647 



Now 2Y+K 3 + H 3 =0 and 2Y'+K 2 ' + H 2 '=0. 

 Hence, from the values just found, we get 



Ox+gY'-fXx^O. 



Also YJ+YT-Xi2i-XiV=0. 



4Y + 2Y'-2X 1 -X 1 ' = 0. 



From these three equations we find Y, Y', Ci in terms of 

 X, and Xx'. 



lhllS \ = Wi X lT Tag A l, 



/ 1 4095 Y i 3999 Y ' 



~ ^1 — 11776 A l + 23552 A l ' 



We want to find the displacements at the load points, i. e. 

 at t==i and t' = ±. 

 We have (from § 9) 



^ = Yt\'d - 1) + K n t + H H . 



Here t = ^, n = 2. If I be the areal moment of inertia for 

 the first portion of the first span, I 2 = ^I. 



3EIW _ 5V ,ip 21 v 15 v , 5 V , 3 Y 



^3 -Si-+4^1- 6 1 Y — til A l+6i Y +32 A l 



— 4^1 + 8 1 64 A l 



_ 4047 Y _ 1935 x ' 



~~ 47104 A l 91208 A l * 



Also - 6E ^ == Y'^(3-0 + K„V + H„'. 



Here t r = \ and ?i = 2. 

 12EIw 



iY'+id-SY'-gXx' + SY'+iXi' 



? 3 



_ in i 3V' 9 Y ' 



— 2^1+8 *■ — 64 A 1 



1935 y p 855 y f 



— "~ 25552 A l ~" 47IU4 A l ' 



Thus we have 



188416= : w=5396X 1 + 1290X 1 ' J 



188416 ? I ?; = 1290X 1 + 1285X 1 '. 



