144 KANSAS UNIVERSITY QUARTERLY. 



Px 



we find V^= '. f'rom 2 (vertical components)=o we find Vn=P 



( 1 ( , 



These vertical reactions are independent of the shape of the rib as^ 

 they do not involve z or y and are the same as for a straight beam. 



Making a section at any point n to the right of the load, consider- 

 ing the left portion of the rib as a free body and denoting the moment 

 by Mr, we have 



Mr=VjZ-P(z--x)-Hy. 



Substituting for V, and y their values in this equation we have 



( z / 4h 



M,-PX j I-^ I __l^-(2l_^,2)H (I) 



Making a section at any point to the left of the load and denoting 

 the moment by Mi, we have 



Mi=VjZ— Hy=P I —^ ( z- l^(zl^z2)H 



(2) 



In equations (i) and (2) H is a function of P and x but not of y or 

 z. To find its value we make use of the equation of the "x displace- 

 ment," viz., /\x= -^ I Myds, ds being the increment of arc, E the 

 EI t/ 



modulus of elasticity and I the moment of inertia. Since this dis- 

 placement is zero we have the equation I Myds=o from which to find 



value of H. 



If the rib is quite flat we may use dz in place of ds. Making this 

 assumption, substituting for M and y their values, and integrating 

 between limits x and o on the left of the load, and 1 and x on the 

 right of the load we have 



r^ ri ( 1— X I r^ 4hH c^ 



i i_x ) ri r^ 4hH n 



dzH-P-j -— j-J ^(Iz2-Z3)dz-Pj ^(z~x)(zl-z2)dz-^-^-j ^ 

 (Iz Z2)3dz=:0. 



Integrating and simplifying this equation we have 



rx— 2l)H Pxl3— ^ =0 (3) 



12 ^ - ' 12 30 ^^-^ 



Solving (3) we have 



, Px ( r x 1 3 r X I 2 I Px 



