CONTINUOUS BEAMS 75 



Now, forming a moment equation for the portion BC, taking center 

 of moments at j5, we have 



whence 

 (73) 



and, eliminating $f between this equation and the expression for 



Z> c , the result is 



Since the deflections at A and C lie on opposite sides of "the tan- 

 gent at j#, we have, from similar triangles, 



Substituting in this relation the values of D A and D c just found, 

 combining like terms, and transposing, we obtain the relation 



(75) 



In this relation M^ M^ M a are the stress couples acting on the 

 beam. The external moments at the supports are equal in amount 

 but opposite in sign to the stress couples. Therefore, calling M^ M z , 

 M s the external moments at the supports, the sign of the expression 

 is changed ; that is, 



(76) JtfA 



which is the required theorem of three moments for a single con- 

 centrated load in each span. 



For a single concentrated load at the center of each span, each 

 k = 1. In this case the theorem becomes 



(77) JfA + 2 M&i + l z ) + M,l z = - 5 (FJ* + pjl). 



o 



If there are a number of concentrated loads in each span, an 

 equation like (76) can be written for each load separately. By 



