of Continuous Girders* 185 



Let m and s denote the moment and shearing-force at any sec- 

 tion ; then from (I.) 



m = M r — S^ + P^— a) for a section between P"| 



and the following support, > . (6) 



?n = M n — S n x for any other section. J 



Also we have 



s = P — S r for a section between P and the r+ lth^j 



support, y . (7) 



s= — S n for any other section. J 



Equations (6) and (7) refer to a concentrated load ; for a uni- 

 form load we have only to put P= £wda, as in the other cases. 



Case II. One end free, the other fastened horizontally. 



tj l 2 lr Is— 2 Is— 1 



2 



A A A A a *i A A 



1 2 3 r T if r + l *-] 



If in (1) to (3) we make 4=0, it is evident that the elastic 

 line passes through two consecutive points, and hence the tan- 

 gent at that point is horizontal ; therefore for this case we have 

 the following rule: — Make 4 = 0, and let s— 1= number of 

 spans ; then formulse (1) to (7) are directly applicable. 



Case III. Both ends fastened horizontally. 



I I, Ic, lr ls—1 Is P 



P A A A «-**! A A * l - 



3 4 r ¥ P r+l «-l 



Let the indices be placed as in the figure ; then in formulae 

 (1) to (7) make Z, = and 7 S =0, and lets — 2= number of spans. 



From the above equations can be easily solved all the exer- 

 cises given in text- books for girders of one span, and also 

 thousands of interesting problems for beams of many spans; 

 (1) to (7) give the moments and shears at every cross section. 

 Making m = in (6), the value of os gives the place of the inflec- 

 tion-points. The deflection at any point x is given by y from 

 (II.) ; and its maximum value is obtained by the usual methods. 

 The following examples have been chosen to illustrate the use 

 of the formulse. 



1. In a girder of four equal spans with free ends, to find the 

 moments at every support due to a single load in the centre of 

 one of the end spans. 



P 



® 



A A A A A 



12 3 4 5 



