CHAP. XII.] METHODS COMBINED. 183 



raent of inertia I, we have for the moment at the centre sup- 

 port B due to any number of weights in both spans, 



a being always measured from the left support. 



Now in this expression the last two terms are precisely the 

 same as for supports on a level ; the influence of the different 

 levels is contained in the first term on the right only. Now by 

 the supposition, <? t and <% must be taken equal to the deflection 

 due to the dead load, and the value of this term will therefore 

 be entirely independent of the live load, which enters only in 

 the last two terms. 



A particular case may perhaps render this plainer. If a 

 girder of two equal spans over three level supports is uni- 

 formly loaded, the reaction at an end support is, as is well 

 known, |ths of the load on one span. 



Now let us take the girder over three supports not on a level, 

 and from our formula above find the reaction at one end due 

 to uniform load when <\ and c^ have the proper values given 

 to them. First the dead load p I over each span causes a de- 



flection at each end of the two cantilevers = -J "s^rt This, 



Hi i 



then, is the value for c t and c 2 in the formula. Now let us 

 take an additional moving load of ml over the whole beam, 

 and with this value of c^ c% find the reaction. We have from 

 our formula 



4 MZ = - 



or M B = \p V \ m Z 2 . 



Now we have by moments, 



R A X Z-tp + )2=+M B ; 



hence, inserting the value of M B above, 



R A = | m I. 



* The&rie tier Trager : Weyrauch. Also Supplement to Chap. XIII., Art. 3. 

 f Supplement to Chap. VII., Art. 13. 



