172 DEFLECTION OF BEAMS 



R 2 c = P 1 m + P 2 n 

 and 



c 

 To find J?j take moments about 3 in (a), and 



(20) 



and from similar triangles in (c) 



R 1 a + R 2 c P i (m + b) P 2 (n d) =o (22) 



Substituting the value of R 2 from (20) in (22), we have 

 R l a = P 1 b P 2 d 



P 1 b P 2 d 



*! = (19) 



a 



and since 



g 



Uniform Load. For a uniform load on the beam the areas of the 

 diagram covered by the uniform load will be used in the place of the 

 ordinates as in Fig. 9 (see discussion on Influence Diagrams, Chapter 

 X). For example in Fig. 10 the reactions are given by the following 

 formulas : 



p (area B 1 area B 2 ) 



a 



area B r 



(24) 



p (area A + area B 2 + area 



R 2 = 



^ (area C 2 - area C l} 



g 



Draw Bridge with Four Supports. To find the reaction at R 2 

 in Fig. n, proceed as follows : With a load represented by the triangle 

 1-2-4, construct a force polygon (not shown) and draw an equi- 



