APPLNDIX.] SUPPORTS OUT OF LEVEL. 385 



Then, <% = 1, <% = 4, c 4 = 15, c 5 56, c e = 209, 

 . ^ _ 540 E I A, ... 906EI/^ lwr _576EIA 3 



1 M2 ~ 209Z 2 ' 3 ~ 209 ? ' M4 ~~209~F"' 



144EIA,, _ _ 



^~ 209? ' 1- 6 ~ ' 



1 3 375 



or, inserting the constants above, and I = b d* = , 



M! = M 6 = 0, M 2 = 5448, M 3 = - 9142, M 4 = 5812, 



M 5 = - 1453 ft. Ibs. 



If all the spans are unloaded. For the reactions necessary 



, to bend it and keep it down to the supports, R! = 272 Ibs., 



Ha = 1002, R 3 = - 1477, R 4 = 1111, R 5 = - 436, Re = 73. 



If the beam weigh 75 Ibs. per foot, what deflection of third 



support will raise the left end from the abutment f 



Ans. R l= wl = A , or Ag = 0.0226 f t.=0.2712 ii. 



It will be observed that a small difference of level has then 

 considerable effect. 



Ex. 5. Two equal spans are uniformly loaded. Sow high 

 must the centre be raised in order that the ends may just 

 touch f 



This is the case of the pivot span with centre support raised. 

 (See Art. 121.) 



The reactions at the ends are zero. At pier, then, Rg = 2 w Z, 

 hence moment at pier M 2 = -J w Z 2 . But the moment when the 

 supports are on level is M 2 = -J w ft, hence w 1? must be due 

 to the elevation of the support. Then from our formulae, 



3 "IP T X ... 74 



Xli 1 /?2 , W L 



which is precisely the same as the deflection of an horizontal 

 beam, fastened at one end, and free at the other (Supplement 

 to Chap. YIL, Art. 13). 



25 



