ART. 1.] SUPPLEMENT TO CHAP. XIIL 243 



SUPPLEMENT TO CHAPTER XIII. 



DEMONSTRATION OF ANALYTICAL FORMULA GIVEN IN TEXT- 



IN the following we shall give the complete development of the general 

 formulae of Art. 155. As these formulae include, as we have seen, all the 

 others as special cases, it is sufficient to show how they are obtained in 

 order to enable the reader to deduce all the others. 



1. Conditions of Equilibrium. In the rth span of a continuous 

 girder, whose length is It (see Fig.), take a point o vertically above the rth 



+ l 



support as the origin of co-ordinates, and the horizontal line o I as the axis 

 of abscissas. At a distance a from the left support pass a vertical section, 

 and between the support and this section let there be a single load P r 

 whose distance from the support is a. 



Now all the exterior forces which act on the girder to the left of the 

 support r we consider as replaced, without disturbing the equilibrium, by 

 a resultant moment M r and a resultant vertical shearing force ST. This 

 moment is equal and opposite to the moment of the internal forces at the 

 section through the support r ; while the vertical force is equal and oppo- 

 site to the shear. 



Not only over the support, but also at every section, the interior forces 

 must hold the exterior ones in equilibrium, and therefore we have the con- 

 ditions : 



1st. The sum (algebraic) of all the horizontal forces must be zero. 



2d. The sum (algebraic) of all the vertical forces must be zero. 



Sd. The sum (algebraic) of the moments of all the forces must be zero. 



Thus, for the section x, we have from the third condition 



S M = M r Srar + Pr (x a) - m = .... (1) 



where m is the moment at the section. From this we have 



m = Mr Sr x + P r (x a) ........ (2) 



If in this we make x = If, m becomes M r +i, and we thus have for the shear 

 just to the right of the left support of the loaded span 



3P L 

 tr 



