Arched Ribs of Uniform Section. 387 



due to such a load, 



dV 



_ = _w' r 4 {i- f .cos</> + ^cosS^}, 



dM 



-wrti 



sin2</> — <£J-. 



(B) 



By means of these two sets of equations, combined with the set 

 given above for the arbitrary constants, all questions concerning 

 circular ribs loaded as described can be solved. It is to be ob- 

 served that (j> is supposed measured from the crown of the rib. 

 The values of F, H, M at any point distant <£ from the crown are 



F = K + F o .cos(£+H o .sin0, 

 U=zpr+-— +H .cos(£— F .sin(£, 



M 



where 



— r 1 Kd(f> + F r sin <f> + H r (1 — cos <f>) -f M ( 



p =w . cos <f> + W 1 . COS 2 (£, 



K= —wr . ^> .cos (£ — iw'r . sin2<p. 



For example, suppose a semicircular rib under a vertical load 

 of uniform horizontal intensity w' but of inconsiderable weight. 

 Also let the rib be hinged at its crown and fixed firmly at its 

 springing; then, since the initial section is at the crown F =0, 



M =0, and H is determined from the equation -jjr — ®> tne 



dn 



value of this coefficient being taken from (B) and the corre- 

 sponding value in the preceding section. Hence we get, 



remembering that <£= ^> 



h o'' 3 {t - 3 } +W> A {! + i-T} =0, 

 whence 



H =JVr. ~ ~ =§w'r approximately. 



And therefore 



F = — jw'r sin 2<f> + § w'r . sin <£, 



H = w'r . cos 2 fy—wW . cos 2<£ + J-w/r . cos <£, 



M = JwV 2 (cos 20 - 1) + §zi>V 2 (l - cos 0), 



