CONTINUOUS BEAMS. 



159 



In place of using this method, which is quite similar to that 

 outlined in the previous article, the equation of the elastic line 

 will be used indirectly as suggested above. 



Since the beam is horizontal at the middle, each arm may be 

 considered like a cantilever carrying a uniform load acting 

 downward, and a load at the end acting upward, such that the 

 end deflection will be zero. (Or the beam may be considered as 

 one simply supported at the ends, with a uniform load and a 

 concentrated load (R 2 ) acting in opposite directions). 



The deflection at the end is 



^77 for the uniform load (Fig. 100). 



for the reaction (Fig. 94). 



and these must be equal. 

 wl* 



Rl = % wl 



Since, by symmetry, the end reactions are the same, the 

 middle reaction R 2 = 2ivl%wl = -:iwl. These results are the 

 same as those given with Fig. 102, and the deduction of the other 

 equations will be exactly the same in both cases. The deduction 

 of some of the equations may be simpler if the origin is taken 

 at A. Each arm is resolvable into a beam supported at its ends 

 of %l span, and a cantilever of length y^l. The maximum mo- 

 ment in the former is %X%wX%Z==Ti8^ 2 , which is the same 

 as that obtained from the general equation when x = %l. In 

 like manner, from the cantilever, the moment at the center sup- 

 port is %ti;ZX W %w*X%* = - y s wl*. 



Fig. 98 shows a case similar to a continuous beam of two 

 equal spans symmetrically loaded, and the equations are de- 

 ducible in a manner similar to the above. The cases of Figs. 99 

 and 103 are the same as one span of a continuous beam of an 

 indefinitely large number of equal spans, similarly loaded, and 

 the equations may be checked by applying the general equation 

 for continuous beams which will now be deduced (97) . 



It should be noted in Fig. 108 that the shear on either side 

 of R 2 is not equal to the reaction, but the reaction is equal to the 

 numerical sum of the shears. The moment is a maximum at three 

 of the points where the shear passes through zero, one of these 

 moments being negative and the other two positive. 



If the middle support were raised, the end reactions would 



