192 STRENGTH OF MATERIALS 



form as for a rib with two hinges, except that its base is shoved 

 down a distance M below the springing line of the rib. Therefore 

 in this case the linear arch is completely determined by the two 

 quantities M and r, the third condition being supplied by the sym- 

 metry of the figure. 



In Fig. 137 let ACF represent the center line of the rib, A'BF f 

 the linear arch, and ADF the equilibrium polygon for the given 

 system of loads. Since the bending moment J/ at any point of the 

 rib is the vertical intercept BC between the linear arch and the 

 center line of the rib, we have 



M = BC = BE- CE' - Ei: . 

 or, since BE = r-DE', M =r-DE' - z - M v 



Substituting this value of M in the first and third of equations (103), 

 they become 



and 



If the expressions under these integral siiins an- evaluated fora num- 

 ber of vertical sections taken at equal distances along the rib, and the 

 results are summed, we obtain the two condition- 



and 



X " u JV * 



r z* ^ 



from which r and M can easily be determined. The linear arch is 

 then constructed by starting from a point at a distance M below tin- 

 left support, and decreasing the ordinates to the equilibrium polygon 

 in the ratio r : 1. 



If the loading is unsymmetrical, the moments at the ends of the 

 rib are not equal. Let M l and M z denote the moments at the left 

 and right ends respectively (Fig. 138). As before, the moment M 

 at any point of the rib is the vertical intercept BC between the liiu-ar 

 arch A f BF' and the center line of the rib ACF. Consequently 



M = BC = BE - CE' - EE f . 



