CONTINUOUS BEAMS 



71 



For a continuous beam bearing a uniform load let A, B, C denote 

 any three consecutive points of support, assumed to be in the same 

 line, and let M^ M Z , M S ; R^ R^ fi & denote the moments and reac- 

 tions at these three points respectively. Also let Z 1? Z 2 denote the 

 lengths of the two spans considered, w l9 w 2 the unit loads on them, 

 and #f, $f the shears on the right and left of R l respectively 

 (Fig. 70), with a similar notation for the other points of support. 



Now consider a portion of the beam cut off by planes just inside 

 the supports at A 

 and (7, as shown in 

 Fig. 71. Then, con- 

 sidering the end B 

 as fixed, the deflec- 

 tion at A from the 



tangent at B consists of three parts : that due to the moment 

 to the shear f considered as a load, and to the uniform load 

 Calling these deflections d^ d^ d^ respectively, we have 



-h 



8 JET 



d = 



"A* 



8JSJ 



(Eq. (50), Art. 39) 

 (Eq. (38), Art. 37) 

 (Eq. (47), Art. 38) 



Hence the total deflection D A of the point A measured from a 

 tangent at the point B is 



MJl S?l* wj* 

 ( 61 ) DA = ~2~#7 + ZEI~ 'SHl' 



To eliminate the shear f, form a moment equation by taking 

 moments about the point B. Then 



T' 



whence 

 (62) 



