ARCHES AND ARCHED RIBS 



191 



necessary to find the partial derivatives of W with respect to M , J^ v 

 and P h respectively, and equate these derivatives to zero. The three 

 conditions obtained in this way are 



(103) T^T = | o^<w = / ^7 ds = 



M 



Mz 



since from either of the above expressions for M we have - - = 1, 

 = x, and - = z. Inserting in these three conditions the value 



h 



of M for the given form of loading, three simultaneous equations are 

 obtained which may be solved for the three unknown quantities B v 

 P k , and M . 



Equations (103) can also be obtained by assuming as our three con- 

 ditions that the horizontal and vertical deflections of the supports are 

 zero, and that the direction of the rib at the ends remains unchanged. 

 The method of obtain- D 



ing equations (103) from 

 these assumptions is 

 simply an extension of 

 that given in Article D / 



152 for the two-hinged 

 arched rib. 



156. Graphical deter- 

 mination of the linear 

 arch for continuous 

 arched rib. The simplest 

 method of applying equa- 

 tions (103) to the deter- 

 mination of the linear 

 arch is by means of a graphical treatment similar to that given in 

 Article 153. 



Consider first the case of symmetrical loading. Then if M denotes 

 tlit- bending moment at either abutment, the linear arch has the same 



E' 



E 



FIG. 137 



